COMPLEX NUMBER. Every Complex Number Can Be Regarded As. Purely imaginary if a = 0. (A) 0 (B) 2i (C) 2i (D) 2

Size: px
Start display at page:

Download "COMPLEX NUMBER. Every Complex Number Can Be Regarded As. Purely imaginary if a = 0. (A) 0 (B) 2i (C) 2i (D) 2"

Transcription

1 J-Mathematics COMPLX NUMBR. DFINITION : Complex umbers are defied as expressios of the form a + ib where a, b R & i = by i.e. = a + ib. a is called real part of (Re ) ad b is called imagiary part of (Im ). very Complex Number Ca Be Regarded As. It is deoted Purely real if b = 0 Purely imagiary if a = 0 Imagiary if b 0 Note : (i) (ii) The set R of real umbers is a proper subset of the Complex Numbers. Hece the Complex Number system is N W I Q R C. Zero is both purely real as well as purely imagiary but ot imagiary. (iii) i = is called the imagiary uit. Also i² = l ; i = i ; i 4 = etc. I geeral i 4 =, i 4+ = i, i 4+ =, i 4+ = i, where I (iv) a b = a b oly if atleast oe of either a or b is o- egative. Illustratio : The value of i 57 + /i 5 is :- (A) 0 (B) i (C) i (D) Solutio : i 57 + /i 5 = i 56. i + i 4.i 4 = i 4 i 4 i = i i i i i i 0 i i As. (A) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65. ARGAND DIAGRAM : Master Argad had doe a systematic study o complex umbers ad represeted every complex umber = x + iy as a set of ordered pair (x, y) o a plae called complex plae (Argad Diagram) cotaiig two perpedicular axes. Horiotal axis is kow as Real axis & vertical axis is kow as Imagiary axis. All complex umbers lyig o the real axis are called as purely real ad those lyig o imagiary axis as purely imagiary.. ALGBR AIC OPR ATIONS : Fudametal operatios with complex umbers : ( a ) Additio (a + bi) + (c + di) = (a + c) + (b + d)i ( b ) Subtractio (a + bi) (c + di) = (a c) + (b d)i ( c ) Multiplicatio (a + bi) (c + di) = (ac bd) + (ad + bc)i a bi ( d ) Divisio a bi. c di ac bd bc ad i c di c di c di c d c d 7 Imagiary axis P() y x Real axis O = x + iy Re()=x, Im()=y

2 J-Mathematics Note : (i) (ii) (iii) The algebraic operatios o complex umbers are similar to those o real umbers treatig i as a polyomial. Iequalities i complex umbers (o-real) are ot defied. There is o validity if we say that complex umber (o-real) is positive or egative. e.g. > 0, 4 + i < + 4i are meaigless. I real umbers, if a + b = 0, the a = 0 = b but i complex umbers, + = 0 does ot imply = = 0. Illustratio : Solutio : i si i si will be purely imagiary, if = (A), I (B), I (C), I (D) oe of these i si will be purely imagiary, if the real part vaishes, i.e., i si 4 si i 8 si = 4 si ( i si ) ( i si ) ( i si ) ( i si ) 4 si 4 si si = 0 = 4 si = 0 (oly if be real) si = ±, I As. (C) Do yourself - : (i) Determie least positive value of for which i i (ii) Fid the value of the sum 5 (i i ), where i =.. QUALITY IN COMPLX NUMBR : Two complex umbers = a + ib & = a + ib are equal if ad oly if their real & imagiary parts are respectively equal. Illustratio : The values of x ad y satisfyig the equatio ( i)x i ( i)y i i are i i (A) x =, y = (B) x =, y = (C) x = 0, y = (D) x =, y = 0 ( i)x i ( i)y i Solutio : i (4 + i) x + (9 7i) y i = 0i i i quatig real ad imagiary parts, we get x 7y = ad 4x + 9y =. Hece x = ad y =. As.(B) 8 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

3 J-Mathematics Illustratio 4 : Fid the square root of i. Solutio : Let 7 4i = a + ib Squarig a b + iab = 7 + 4i Compare real & imagiary parts a b = 7 & ab = 4 By solvig these two equatios We get a = ±4, b = ± 7 4i = ±(4 + i) Illustratio 5 : If x 5 4, fid the value of x 4 + 9x + 5x x + 4. Solutio : We have, x = x + 5 = 4i (x + 5) = 6i x + 0x + 5 = 6 x + 0x + 4 = 0 Now, x 4 + 9x + 5x x + 4 x (x + 0x + 4) x(x + 0x + 4) + 4(x + 0x + 4) 60 x (0) x(0) + 4(0) A s. Do yourself - : (i) Fid the value of x + 7x x + 6, where x = + i. (ii) If a + ib = c i c i, where c is a real umber, the prove that : b c a + b = ad a c. (iii) Fid square root of 5 8i NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 4. THR IMPORTANT TRMS : CONJUGAT/MODULUS/ARGUMNT : ( a ) CONJUGAT COMPLX : If = a + ib the its cojugate complex is obtaied by chagig the sig of its imagiary part & is deoted by. i.e. = a ib. Note that : (i) (ii) (iii) + = Re() = i Im() = a² + b², which is purely real (iv) If is purely real, the = 0 (v) If is purely imagiary, the + = 0 (vi) If lies i the st quadrat, the lies i the 4 th quadrat ad ( b ) Modulus : lies i the d quadrat. 9 Im If P deotes complex umber = x + iy, the the legth OP is called modulus of complex umber. It is deoted by. OP = = x y Geometrically represets the distace of poit P from origi. ( 0) Note : Ulike real umbers, = is ot correct. if 0 if 0 Re

4 J-Mathematics ( c ) Argumet or Amplitude : Imagiary If P deotes complex umber = x + iy ad if OP makes a agle axis with real axis, the is called oe of the argumets of. = ta y x Note : (i) (ii) (agle made by OP with positive real axis) O P(x, y) Real axis Argumet of a complex umber is a may valued fuctio. If is the argumet of a complex umber, the + ; I will also be the argumet of that complex umber. Ay two argumets of a complex umber differ by The uique value of such that < is called Amplitude (pricipal value of the argumet). (iii) Pricipal argumet of a complex umber = x + iy ca be foud out usig method give below : (iv) (a) Fid = (b) ta y x such that Im 0,. Use give figure to fid out the pricipal argumet accordig as the poit lies i respective quadrat. Uless otherwise stated, amp implies pricipal value of the argumet. Re (v) (vi) y The uique value of = ta x such that 0 is called least positive argumet. If = 0, arg() is ot defied (vii) If is real & egative, arg() =. (viii) If is real & positive, arg() = 0 (ix) (x) Illustratio 6 : If If, lies o the positive side of imagiary axis., lies o the egative side of imagiary axis. By specifyig the modulus & argumet a complex umber is defied completely. Argumet impart directio & modulus impart distace from origi. For the complex umber 0 + 0i the argumet is ot defied ad this is the oly complex umber which is give by its modulus oly. Fid the modulus, argumet, pricipal value of argumet, least positive argumet of complex umbers (a) + i (b) + i (c) i (d) i Solutio : (a) For = + i ( ) arg () = +, I Least positive argumet is 0 y 60 (, ) If the poit is lyig i first or secod quadrat the amp() is take i aticlockwise directio. I this case amp() = x NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

5 J-Mathematics (b) For = + i = (, ) y arg () = +, I Least positive argumet = amp() = 60 0 x (c) For = i y = arg () =, I 5 / / x Least positive argumet = 5 (, ) If the poit lies i third or fourth quadrat the cosider amp() i clockwise directio. I this case amp() = (d) For = i = arg () =, I Least positive argumet = 4 amp() = 60 (, ) y 4 / / x NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 Illustratio 7 : Fid modulus ad argumet for = si + i cos, (0,) Solutio : ( si ) (cos ) si cos si Case (i) For 0,, will lie i I quadrat. amp () = arg cos ta amp () = ta si ta ta 4 Sice, 4 4 amp () = Case (ii) at : 0 0i = 0 amp () is ot defied. 4, cos si cos si cos si cos si cos si ta

6 J-Mathematics Case (iii) Case (iv) Case (v) Sice For,, will lie i IV quadrat so amp () = ta ta 4 Sice, 4 amp () = 4 4 at : = + 0i = amp () = 0 For, will lie i I quadrat arg () = ta ta 5, 4 4 4, = si cos arg = = 4, = si cos 4 Do yourself - : Fid the modulus ad amplitude of followig complex umbers : (i) i (ii) i (iii) i (iv) i i ( v ) 6 i 5 i 5. RPRSNTATION OF A COMPLX NUMBR IN VARIOUS FORMS : ( a ) Cartesia Form (Geometrical Represetatio) : very complex umber = x + i y ca be represeted by a poit o the cartesia plae kow as complex plae by the ordered pair (x, y). There exists a oe-oe correspodece betwee the poits of the plae ad the members of the set of complex umbers. For = x + iy; y x y ; x iy ad ta x Note : (i) Distace betwee the two complex umbers & is give by. (ii) 0 = r, represets a circle, whose cetre is 0 ad radius is r. Illustratio 8 : Fid the locus of : (a) + + = 4 (b) Re( ) = 0 Solutio : (a) Let = x + iy ( x + iy ) + ( x + iy + ) = 4 (x ) + y + (x + ) + y = 4 x x + + y + x + x + + y = 4 x + y = Above represets a circle o complex plae with ceter at origi ad radius uity. O Imagiary axis P(x, y) Real axis NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

7 (b) Let = x + iy = x y + xyi Re( ) = 0 x y = 0 y = ± x Thus Re( ) = 0 represets a pair of straight lies passig through origi. Illustratio 9 : If is a complex umber such that = ( ), the (A) is purely real (C) either is purely real or purely imagiary Solutio : Let = x + iy, the its cojugate x iy Give that = (B) is purely imagiary (D) oe of these ( ) x y + ixy = x y ixy 4ixy = 0 If x 0 the y = 0 ad if y 0 the x = 0. J-Mathematics As. (C) Illustratio 0 : Amog the complex umber which satisfies 5i 5, fid the complex umbers havig Solutio : (a) least positive argumet (c) least modulus The complex umbers satisfyig the coditio 5i 5 (b) maximum positive argumet (d) maximum modulus are represeted by the poits iside ad o the circle of radius 5 ad cetre at the poit C(0, 5). The complex umber havig least positive argumet ad maximum positive argumets i this regio are the poits of cotact of tagets draw from origi to the circle Here = least positive argumet ad = maximum positive argumet I OCP, OP OC CP ad OP 0 4 si OC ta ta Q C 5i O D40i P N origi Taget from Thus, complex umber at P has modulus 0 ad argumet ta 4 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 4 p 0 cos i si 0 i 5 5 Do yourself - 4 : (i) 6i p Similarly Q = + 6i From the figure, is the poit with least modulus ad D is the poit with maximum modulus. Hece, O OC C 5i 5i 0i ad D OD OC CD 5i 5i 40i Fid the distace betwee two complex umbers = + i & = 7 9i o the complex plae. (ii) Fid the locus of i =. (iii) If is a complex umber, the + represets - (A) a circle (B) a straight lie (C) a hyperbola (D) a ellipse

8 J-Mathematics ( c ) Trigoometric / Polar Repre setat io : = r (cos + i si ) where = r ; arg = ; = r (cos i si ) Note : cos + i si is also writte as CiS uler' s formula : The formula e ix = cosx + i si x is called uler's formula. It was itroduced by uler i 748, ad is used as a method of expressig complex umbers. Also cos x = e ix e ix & si x = ( d ) xpoetial Represetatio : e ix e i ix are kow as uler's idetities. Let be a complex umber such that = r & arg =, the = r.e i Illustratio : xpress the followig complex umbers i polar ad expoetial form : (i) i i (ii) i cos i si Solutio : (i) Let i i i i i i i ( ) ta ta 4 4 Re() < 0 ad Im() > 0 lies i secod quadrat. = arg () = = 4 4 Hece Polar form is = cos i si 4 4 (ii) / 4 ad expoetial form is e i i (i ) Let = cos i si i ( i ) (i ) ( i ) i ( i ) ( i ) Re() > 0 ad Im() > 0 lies i first quadrat. ( ) ta ta 5 5 Hece Polar form is cos i si 5 / ad expoetial form is e 4 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

9 Illustratio : If x = cos i si the x x x... is equal to - (A) (B) (C) 0 (D) Solutio : x = cos i si = i e x x x... i i i = e.e e = e = i cos... + i si... = J-Mathematics / as... / As. (A) Do yourself - 5 : xpress the followig complex umber i polar form ad expoetial form : ( 7i) (i) + i (ii) i (iii) ( i) 6. IMPORTANT PROPRTIS OF CONJUGAT : ( a ) + = Re () ( b ) = i Im () (c) ( ) = ( d ) = + (e) = ( f ) =.. I geeral (iv) ( cos + isi), (0) ( g ) = ; 0 (h ) If f( + i) = x + iy f( i) = x iy 7. IMPORTANT PROPRTIS OF MODULUS : ( a ) 0 ( b ) Re () (c) Im () ( d ) = = (e) = NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 ( f ) =.. I geeral (g) = (h ) =, I, 0 (i) + = + + Re ( j ) + = + + cos( ), where, are arg( ), arg( ) respectively. ( k ) + = ( l ) + + [Triagle Iequality] ( m ) + [Triagle Iequality] 5

10 J-Mathematics 8. IMPORTANT PROPRTIS OF AMPLITUD : ( a ) amp (. ) = amp + amp + k k I ( b ) amp = amp amp + k k I ( c ) amp( ) = amp() + k;,k I where proper value of k must be chose so that RHS lies i (, ]. Illustratio : Fid amp ad if ( 4i)( i)( i). ( i)(4 i)(i) Solutio : amp = amp ( 4i) amp( i) amp( i) amp( i) amp(4 i) amp(i) k where k I ad k chose so that amp lies i (,] amp ta ta k amp = Also, amp = Aliter 4 4 ta cot + k amp k 6 [at k = ] A s. 4i i i i4 ii 5 5 ( 4i)( i) i i4 ii Hece =, amp() =. I l l u s t r a t i o 4 : If i, the locus of is - i Solutio : 4i i i i 4 i i i i i = 4 (A) x-axis (B) y-axis (C) x = (D) y = We have, i x i y i x i y x i y x y x y x i y Illustratio 5 : If + = + the (A) ero or purely imagiary (C) purely real is - A s. 4y 0; y 0, which is x-axis As. (A) (B) purely imagiary (D) oe of these NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

11 J-Mathematics cos i si, r Solutio : Here let = r = r cos i si, r r cos r cos i r si r si ( + ) = r r r r cos( ) = if cos( ) 0 = amp( ) amp( ) = amp is purely imagiary As. (B) Illustratio 6 : ad are two complex umbers such that Solutio : Here is ot uimodular. Fid. = is uimodular (whose modulus is oe), while But (give) = 4 Hece, =. I l l u s t r a t i o 7 : The locus of the complex umber i argad plae satisfyig the iequality NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 4 log / where (A) a circle (B) iterior of a circle (C) exterior of a circle (D) oe of these 4 Solutio : We have, log / log / I l l u s t r a t i o 8 : 4 loga x is a decreasig fuctio if a is - 8 as > / 0 which is exterior of a circle. If 4 =, the the greatest value of is - (A) + (B) + (C) + (D) 5 + As. (C) 7

12 J-Mathematics Solutio : We have = Therefore, the greatest value of is 5 +. As. (D) Illustratio 9 : Shaded regio is give by - (A) + 6, 0 arg() 6 (B) + 6, 0 arg() C(+ i ) (C) + 6, 0 arg() (D) Noe of these A 0 B(4) Solutio : Note that AB = 6 ad + i = + + i = + 6 i = + 6 cos i si BAC = Thus, shaded regio is give by + 6 ad 0 arg ( + ) As. (C) Do yourself - 6 : (i) The iequality 4 < represets regio give by - (A) Re() > 0 (B) Re() < 0 (C) Re() > (D) oe (ii) If = re i, the the value of e i is equal to - (A) e rcos (B) e rcos (C) e rsi (D) e rsi 9. SCTION FORMUL A AND COORDINATS OF ORTHOCNTR, CNTROID, CIRCUMCNTR, INCNTR OF A TRIANGL : m If & are two complex umbers the the complex umber = divides the joi of m & i the ratio m :. Note : (i) If a, b, c are three real umbers such that a + b + c = 0 ; where a + b + c = 0 ad a,b,c are ot all simultaeously ero, the the complex umbers, & are colliear. (ii) If the vertices A, B, C of a triagle represet the complex umbers,, respectively, the : Cetroid of the ABC = Orthocetre of the ABC = a sec A b sec B c sec C a sec A b sec B c sec C Icetre of the ABC = (a b c ) (a b c) 8 or ta A ta B ta C ta A ta B ta C Circumcetre of the ABC = ( si A si B si C) (si A si B si C) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

13 J-Mathematics 0. VCTORIAL RPRSNTATION OF A COMPLX NUMBR : y ( a ) I complex umber every poit ca be represeted i terms of P() positio vector. If the poit P represets the complex umber the, OP = & OP = O x ( b ) If P( ) & Q( ) be two complex umbers o argad plae the y P( ) Q( ) PQ represets complex umber. O x Note : (i) If OP = = r e i the OQ of uequal magitude the = = r e i ( + ) =. e i. If OP ^ ^ i OP e OQ i.e. ad OQ e i are y O r r Q( ) P() x (ii) I geeral, if,, be the three vertices of ABC the y C( ) e i. Here (iii) Note that the locus of satisfyig arg arg. is: A( ) B( ) x Case (a) 0 < < / Locus is major arc of circle as show excludig & NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 Case (b) (iv) (v) Locus is mior arc of circle as show excludig & If A, B, C & D are four poits represetig the complex umbers,, & 4 the AB CD if AB CD if 4 4 is purely imagiary. is purely real ; If,, are the vertices of a equilateral triagle where 0 is its circumcetre the () + + = 0 () + + = 0 9

14 J-Mathematics Illustratio Solutio : 0 : Complex umbers,, are the vertices A, B, C respectively of a isosceles right agled triagle with right agle at C. Show that ( ) = ( )( ). I the isosceles triagle ABC, AC = BC ad BCAC. It meas that AC is rotated through agle / to occupy the positio BC. i / Hece we have, e i = +i( ) B( ) = C( ) A( ) Illustratio : If the vertices of a square ABCD are,, & 4 the fid & 4 i terms of &. Solutio : Usig vector rotatio at agle A e i 4 AC ad AB A( ) 4 D( ) 4 Also AC = AB B( ) C( ) = cos i si 4 4 = ( ) ( + i) = + ( ) ( + i) Similarly 4 = + ( + i)( ) Illustratio : Plot the regio represeted by arg i the Argad plae. Solutio : Let us take arg =, clearly lies o the mior arc of the circle passig through (, 0) ad (, 0). Similarly, arg = meas that '' is lyig o the major arc of the circle passig through (, 0) ad (, 0). Now if we take ay poit i the regio icluded betwee two arcs say P ( ) we get arg Thus 40 (,0) (,0) / arg represets the shaded regio (excludig poits (, 0) ad (, 0)). NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

15 Do yourself - 7 : (i) (ii) J-Mathematics A complex umber = + 4i is rotated about aother fixed complex umber = + i i aticlockwise directio by 45 agle. Fid the complex umber represeted by ew positio of i argad plae. If A, B, C are three poits i argad plae represetig the complex umber,, such that =, where R, the fid the distace of poit A from the lie joiig poits B ad C. (iii) If A( ), B( ), C( ) are vertices of ABC i which ABC = 4 AB ad BC terms of ad. (iv) ( v ) If arg, the fid i If a & b are real umbers betwee 0 ad such that the poits = a + i, = + bi ad = 0 form a equilateral triagle the a ad b are equal to :- (A) a = b = / (B) a = b = (C) a = b = + (D) a = b =, fid locus of. 4. D' MOIVR S THORM : The value of (cos + isi) is cos + isi if '' is iteger & it is oe of the values of (cos + isi) if is a ratioal umber of the form p/q, where p & q are co-prime. Note : Cotiued product of the roots of a complex quatity should be determied by usig theory of equatios. Illustratio : If cos + cos + cos = 0 ad also si + si + si = 0, the prove that (a) cos + cos + cos = si + si + si = 0 (b) si + si + si = si() (c) cos + cos + cos = cos() Solutio : Let = cos + i si, = cos+ isi & = cos + isi. + + = (cos + cos + cos) + i(si + si + si) = 0 + i. 0 = 0... (i) (a) Also cos i si cos i si NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 cos i si, cos i si = (cos + cos + cos) i(si + si + si)... (ii) = 0 i. 0 = 0 Now = 0 = 0. 0 = 0 {usig (i) ad (ii)} or cos i si (cos i si ) cos i si 0 or cos + isi + cos + isi + cos + isi = 0 + i.0 quatig real ad imagiary parts o both sides, cos + cos + cos = 0 ad si + si + si = 0 4

16 J-Mathematics Do yourself - 8 : (b) If + + = 0 the (cos + isi) + (cos + isi) + (cos + isi) = (cos + isi) (cos + isi) (cos + isi) or cos + isi + cos + isi + cos + isi = {cos() + isi()} quatig imagiary parts o both sides, si + si + si = si( ) (c) quatig real parts o both sides, cos + cos + cos = cos( ) r r (i) If r cos i si, r 0,,,4,..., the is equal to - (A) (B) 0 (C) (D) oe of these (ii) If (x ) 4 6 = 0, the the sum of oreal complex values of x is - (A) (B) 0 (C) 4 (D) oe of these (iii) If ( i), Z, the is a multiple of - (A) 6 (B) 0 (C) 9 (D). CUB ROOT OF UNITY : ( a ) The cube roots of uity are, i ( ), 4 i ( ) ( b ) If is oe of the imagiary cube roots of uity the + + ² = 0. I geeral + r + r = 0 ; where r I but is ot the multiple of & + r + r = if r = ; I ( c ) I polar form the cube roots of uity are : Im = cos 0 + i si 0 ; = cos + i si, = cos i si. ( d ) The three cube roots of uity whe plotted o the argad plae costitute the vertices of a equilateral triagle. (e) The followig factorisatio should be remembered : (a, b, c R & is the cube root of uity) a b = (a b) (a b) (a ²b) ; x + x + = (x ) (x ) ; a + b = (a + b) (a + b) (a + b) ; a + b + c abc = (a + b + c) (a + b + ²c) (a + ²b + c) Illustratio 4 : If & are imagiary cube roots of uity the + is equal to - Solutio : (A) cos cos i si cos i si cos i si = (B) cos + cos i si (C) i si O / (D) i si cos i si cos i si = cos As. (A) Re NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

17 J-Mathematics Illustratio 5 : If are roots of x x + x + 7 = 0 (ad is imagiary cube root of uity), the fid the value of. Solutio : We have x x + x + 7 = 0 (x ) + 8 = 0 (x ) = ( ) x x / =,, (cube roots of uity) x =, Here =, =, = =, =, = The = = Therefore =. As. Do yourself - 9 : (i) If is a imagiary cube root of uity, the ( + ) equals : - (A) (B) 4 (C) (D) 4 (ii) If is a o real cube root of uity, the the expressio ( )( )( + 4 )( + 8 ) is equal to : - (A) 0 (B) (C) (D). th ROOTS OF UNITY : If,,,.... are the, th root of uity the : ( a ) They are i G.P. with commo ratio e i(/) ( b ) Their argumets are i A.P. with commo differece ( c ) The poits represeted by, th roots of uity are located at the vertices of a regular polygo of sides iscribed i a uit circle havig ceter at origi, oe vertex beig o positive real axis. ( )A / / / A ( ) A () NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 p ( d ) + p + p p = 0 if p is ot a itegral multiple of = if p is a itegral multiple of (e) ( ) ( )... ( ) = ( f ) ( + ) ( + )... ( + ) = 0 if is eve ad = if is odd. ( g ) = or accordig as is odd or eve. 6 k k Illustratio 6: Fid the value si cos 7 7 Solutio : k k k si cos = k k = 6 k k si cos k k 0 (Sum of imagiary part of seve seveth roots of uity) k 0 6 (Sum of real part of seve seveth roots of uity) + = = k 0 A( ) 4

18 J-Mathematics 4. TH SUM OF TH FOLLOWING SRIS SHOULD B RMMBRD : ( a ) cos + cos + cos cos = ( b ) si + si + si si = Note : si / si / si / si / si cos If = (/) the the sum of the above series vaishes. 5. STRAIGHT LINS & CIRCLS IN TRMS OF COMPLX NUMBRS : y ( a ) amp( ) = is a ray emaatig from the complex poit ad iclied at a agle to the x axis. O x y (a) (b) ( b ) a = b is the perpedicular bisector of the segmet joiig a & b. O x = + t ( ) where t is a parameter. ( c ) The equatio of a lie joiig & is give by ; y ( d ) = ( + it) where t is a real parameter, is a lie through the poit & O y x (e) perpedicular to. The equatio of a lie passig through & ca be expressed i the determiat form as O x = 0. This is also the coditio for three complex umbers to be colliear. ( f ) Complex equatio of a straight lie through two give poits & ca be writte as = 0, which o maipulatig takes the form as r = 0 where r is real ad is a o ero complex costat. ( g ) The equatio of circle havig cetre 0 & radius is : 0 = or ² = 0 which is of the form r = 0, r is real cetre = & radius = r. Circle will be real if r 0. (h ) arg = ± or ( ) ( ) + ( ) ( ) = 0 this equatio represets the circle described o the lie segmet joiig & as diameter. (i) Coditio for four give poits,, & 4 to be cocyclic is, the umber is real. Hece the equatio of a circle through o colliear poits, & ca be take as is real = () NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

19 Miscellaeous Illustratio : J-Mathematics Illustratio 7 : If is a poit o the Argad plae such that =, the is equal to - (A) ta (arg ) (B) cot (arg ) (C) i ta (arg ) (D) oe of these Solutio : Sice =, Illustratio Solutio : let cos i si The, cos i si si i si cos i si cos i si ad cos i si cos i si cos cos cos i si... (i)... (ii) From (i) ad (ii), we get i ta i ta arg arg from ii As. (C) 8 : Let a be a complex umber such that a < ad,,..., be the vertices of a polygo such that k = + a + a +... a k, the show that vertices of the polygo lie withi the circle a a. We have, a a... a k k a k a a k a a k k a k a a a a Vertices of the polygo,,..., lie withi the circle Illustratio 9 : If ad are two complex umbers ad C > 0, the prove that + ( + C) + ( + C ) Solutio : We have to prove that : + ( + C) + ( + C ) i.e. + + C + ( +C ) a a or C C NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 or or C 0 C (usig Re C 0 C ) which is always true. Illustratio 0 : If [/6, /], i =,,, 4, 5 ad 4 cos + cos + cos + cos 4 + cos 5 =, the show that > 4 Solutio : Give that or cos. cos. cos. cos. cos cos. cos. cos. cos. cos cos. cos. cos. cos. cos i / 6, / cos i

20 J-Mathematics < 4 > 4 Illustratio : If,, are complex umbers such that,, lie o a circle passig through the origi., show that the poits represeted by Solutio : We have, O arg arg arg arg or arg = Thus the sum of a pair of opposite agle of a quadrilateral is 80. Hece, the poits 0,, ad are the vertices of a cyclic quadrilateral i.e. lie o a circle. Illustratio : Two give poits P & Q are the reflectio poits w.r.t. a give straight lie P if the give lie is the right bisector of the segmet PQ. Prove that the two poits deoted by the complex umbers & will be the reflectio poits for the straight lie r 0 if ad oly if ; Solutio : r 0, where r is real ad is o ero complex costat. Q Let P( ) is the reflectio poit of Q( ) the the perpedicular bisector of & must be the lie r 0... (i) Now perpedicular bisector of & is, or ( ) ( cacels o either side) 0... (ii) or Comparig (i) & (ii) r... (iii) r... (v) Multiplyig (iii) by ; (iv) by ad addig r (iv) Note that we could also multiply (iii) by & (iv) by & add to get the same result. NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

21 J-Mathematics Hece r 0 Agai, let r 0 is true w.r.t. the lie r 0. Subtractig ( ) 0 or or Hece '' lies o the perpedicular bisector of jois of &. : (i) = 4 (ii) 0 : (i) 7 + 4i (iii) ±( 4i) : (i) = 4; amp() = (iv) ANSWRS FOR DO YOURSLF (ii) ; amp() ( v ) ; amp() 4 5 ; amp() (iii) ; amp() 6 4 : (i) uits (ii) locus is a circle o complex plae with ceter at (,) ad radius uit. (iii) C NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 5 : (i) (iii) cos i si ; e 4 4 cos i si ; e : (i) C (ii) D i 4 i 4 (ii) (iv) 4 4 cos i si ; e i 4 si cos i si ; si e 7 : (i) ( )i (ii) 0 (iii) = + i( ) (iv) B ( v ) Locus is all the poits o the major arc of circle as show excludig poits &. Im /4 O 8 : (i) C (ii) A (iii) D 9 : (i) D (ii) B c(0,) Re i 47

22 J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). The value of the sum i i, where i, equals [J 98] (A) i (B) i (C) i (D) 0. The sequece S = i + i + i +... upto 00 terms simplifies to where i = - (A) 50( i) (B) 5i (C) 5( + i) (D) 00( i). Let i. The product of the real part of the roots of = 5 5i is - (A) 5 (B) 6 (C) 5 (D) 5 4. If = a i, a 0 ad =, b 0 are such that the - bi (A) a =, b = (B) a =, b = (C) a =, b = (D) a =, b = 5. The iequality 4 < represets the followig regio - (A) Re() > 0 (B) Re() < 0 (C) Re() > (D) oe of these 6. If ( + i) ( + i) ( + i)... ( + i) = + i the ( + ) = (A) i (B) (C) + (D) oe of these 7. I the quadratic equatio x + (p +iq) x + i = 0, p & q are real. If the sum of the squares of the roots is 8 the : (A) p =, q = (B) p =, q = (C) p =, q = or p =, q = (D) p =, q = 8. The curve represeted by Re( ) = 4 is - (A) a parabola (C) a circle 9. Real part of i e e is - 48 (B) a ellipse (D) a rectagular hyperbola (A) e cos [cos (si )] (B) e cos [cos (cos )] (C) e si [si (cos )] (D) e si [si (si )] 0. Let ad are two o-ero complex umbers such that = ad arg + arg =, the equal to - (A) (B) (C) (D). Number of values of x (real or complex) simultaeously satisfyig the system of equatios = 0 ad = 0 is - (A) (B) (C) (D) 4. If =, =, = ad = the the value of + + is equal to- (A) (B) (C) 4 (D) 6. A poit moves o the curve 4 i = i a argad plae. The maximum ad miimum values of are - (A), (B) 6, 5 (C) 4, (D) 7, 4. The set of poits o the complex plae such that + + is real ad positive (where = x + iy, x, y R (A) Complete real axis oly (B) Complete real axis or all poits o the lie x + = 0 (C) Complete real axis or a lie segmet joiig poits, &, (D) Complete real axis or set of poits lyig iside the rectagle formed by the lies. x + = 0 ; x = 0 ; y 0 & y 0 excludig both. ) is- NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

23 J-Mathematics 7 5. If is a imagiary cube root of uity, the ( ) equals [J 98] (A) 8 (B) 8 (C) 8 (D) 8 6. If i, the i i 65 is equal to : [J 99] (A) i (B) i (C) i (D) i 7. The set of poits o a Argad diagram which satisfy both 4 & Arg are lyig o - (A) a circle & a lie (B) a radius of a circle (C) a sector of a circle (D) a ifiite part lie 8. If Arg ( i) = 4, the the locus of is - (,) (,) y y (A) (B) (C) x (D) x (, ) 9. The origi ad the roots of the equatio + p + q = 0 form a equilateral triagle if - (, ) (A) p = q (B) p = q (C) p = q (D) q = p 0. Poits & are adjacet vertices of a regular octago. The vertex adjacet to ( ) ca be represeted by - ( i)( ) (B) ( i)( ) (A) ( i)( ) (D) oe of these (C) i i i i is equal to - NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 (A) (B) (C) (D) oe of these. If ad are two o-ero complex umbers such that =, ad Arg () Arg() = /, the is equal to - (A) (B) (C) i (D) i SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS). For two complex umbers ad : (a b )(c d ) (c d )(a b ) if (a, b, c, d R) - (A) a c b d (B) a b (C) (D) arg( d c ) = arg( ) 4. Which of the followig, locii of o the complex plae represets a pair of straight lies? (A) Re( ) = 0 (B) Im( ) = 0 (C) + = 0 (D) = i 5. If the complex umbers,, represets vertices of a equilateral triagle such that = =, the which of followig is correct? (A) (B) Re( + + ) = 0 (C) Im( + + ) = 0 (D) + + = 0 x i 6. If S be the set of real values of x satisfyig the iequality log 0, the S cotais - (A) [, ) (B) (, ] (C) [, ] (D) [, ] 49

24 J-Mathematics 7. If amp ( ) = 0 ad = =, the :- (A) + = 0 (B) = (C) = (D) oe of these 8. If the vertices of a equilateral triagle are situated at =0, =, =, the which of the followig is/are true - (A) = (B) = (C) + = + (D) arg arg = / 9. Value(s) of ( i) / is/are - (A) i i i i (B) (C) (D) 0. If cetre of square ABCD is at =0. If affix of vertex A is, cetroid of triagle ABC is/are - (A) (C) (cos + i si ) (B) 4 cos i si cos i si (D) cos i si. If is a imagiary cube root of uity, the a root of equatio x x x = 0, ca be :- (A) x = (B) x = (C) x = (D) x = 0 CHCK YOUR GRASP ANSWR KY X R CI S - Que A s. B A B B D C C D A D Que A s. A A D B D C C A C B Que A s. A D A, D A, B B,C,D A, B B. C A,B,D A,C C, D Que. A s. D 50 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

25 J-Mathematics XRCIS - 0 BRAIN TASRS SLCT TH CORRCT ALTRNATIVS (ON OR MOR THAN ON CORRCT ANSWRS). O the argad plae, let, & =. The the correct statemet is - (A) moves o the circle, cetre at (,0) ad radius (B) & describe the same locus (C) & move o differet circles (D) moves o a circle cocetric with =. The value of i + i, for i ad I is - (A) ( i) ( i) (B) ( i) ( i) (C) ( i) ( i) (D) ( i) ( i). The commo roots of the equatios + ( + i) + ( + i) + i = 0, (where i = ) ad = 0 are - (where deotes the complex cube root of uity) (A) (B) (C) (D) If x r CiS r for r ; r, N the - (A) Lim Re x r r (B) Lim Re x r 0 r (C) Lim Im x r r (D) Lim Im x r 0 5. Let, be two complex umbers represeted by poits o the circle = ad = respectively, the - (A) max + = 4 (B) mi = (C) (D) oe of these 6. If, be ay two complex umbers such that, the which of the followig may be true - r i i (A) (B) (C) e, R (D) e, 7. Let, ad + represet three vertices of ABC, where is cube root uity, the - R NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 (A) cetroid of ABC is ( ) (B) orthoceter of ABC is ( ) (C) ABC is a obtuse agled triagle (D) ABC is a acute agled triagle 8. Which of the followig complex umbers lies alog the agle bisectors of the lie - L : = ( + ) + i( + 4) L : = ( + ) + i( 4) (A) 5 i (B) + 5i (C) i 5 (D) 5 i 9. Let ad are two complex umbers such that, ad + i = i =, the equals - (A) or i (B) i or i (C) or (D) i or 0. If g(x) ad h(x) are two polyomials such that the polyomial P(x) = g(x ) + xh(x ) is divisible by x + x +, the - (A) g() = h() = 0 (B) g() = h() 0 (C) g() = h() (D) g() + h() = 0 BRAIN TASRS ANSWR KY X R CI S - Que A s. A,B,D B, D B, C A, D A,B,C A,B,C,D A,C A,C C A,C,D 5

26 J-Mathematics XRCIS - 0 MISCLLANOUS TYP QUSTIONS MATCH TH COLUMN Followig questio cotais statemets give i two colums, which have to be matched. The statemets i Colum-I are labelled as A, B, C ad D while the statemets i Colum-II are labelled as p, q, r ad s. Ay give statemet i Colum-I ca have correct matchig with ON statemet i Colum-II.. Colum-I Colum-I (A) If be the complex umber such that the miimum value of is ta 8 (p) 0 (B) = & + 0 the is equal to (q) (C) If 8i i = 0 the = (r) (D) If,,, 4 are the roots of equatio (s) = 0, the 4 ( i + ) is i Followig questio cotais statemets give i two colums, which have to be matched. The statemets i Colum-I are labelled as A, B, C ad D while the statemets i Colum-II are labelled as p, q, r ad s. Ay give statemet i Colum-I ca have correct matchig with ON OR MOR statemet(s) i Colum-II.. Match the figure i colum-i with correspodig expressio - Colum-I Colum-I (A) 4 two parallel lies (p) 4 4 = 0 4 (B) two perpedicular lies (q) (C) 4 a parallelogram (r) (D) (s) + = + 4 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

27 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 ASSRTION & RASON These questios cotais, Statemet I (assertio) ad Statemet II (reaso). 7 + = ( + ) cos 7 cos 7 5 J-Mathematics (A) Statemet-I is true, Statemet-II is true ; Statemet-II is correct explaatio for Statemet-I. (B) Statemet-I is true, Statemet-II is true ; Statemet-II is NOT a correct explaatio for statemet-i. (C) Statemet-I is true, Statemet-II is false. (D) Statemet-I is false, Statemet-II is true.. Statemet-I : There are exactly two complex umbers which satisfy the complex equatios 4 5i = 4 ad Arg ( 4i) = 4 simultaeously. B e c a u s e Statemet-II : A lie cuts the circle i atmost two poits. (A) A (B) B (C) C (D) D. Let,, satisfy 0 Statemet : arg arg. 0 ad 0 =. Cosider least positive argumets wherever required. a d Statemet :,, satisfy 0 =. (A) A (B) B (C) C (D) D. Statemet-I : If = i + i + i i, the,, & forms the vertices of square o argad plae. B e c a u s e Statemet-II :,,, are situated at the same distace from the origi o argad plae. (A) A (B) B (C) C (D) D 4. Statemet-I : If = 9 + 5i ad = + 5i ad if arg B e c a u s e 4 Statemet-II : If lies o circle havig & as diameter the arg the 6 8i =. 4 (A) A (B) B (C) C (D) D 5. Statemet- : Let,, be three complex umbers such that + = + = + ad = 0, the,, will represet vertices of a equilateral triagle o the complex plae. a d Statemet- :,, represet vertices of a equilateral triagle if. (A) A (B) B (C) C (D) D COMPRHNSION BASD QUSTIONS Comprehesio # : Let be ay complex umber. To factorise the expressio of the form, we cosider the equatio =. This equatio is solved usig De moiver's theorem. Let,,,... be the roots of this equatio, the = ( )( )( )...( ) This method ca be geeralised to factorie ay expressio of the form k. 6 for example, 7 m + = C is m This ca be further simplified as 5 cos... (i) 7

28 J-Mathematics These factorisatios are useful i provig differet trigoometric idetities e.g. i eqautio (i) if we put = i, the equatio (i) becomes 5 ( i) (i ) i cos i cos i cos i.e. 5 cos cos cos O the basis of above iformatio, aswer the followig questios :. If the expressio 5 ca be factorised ito liear ad quadratic factors over real coefficiets as 5 ( p 4)( q 4), where p > q, the the value of p q - (A) 8 (B) 4 (C) 4 (D) 8. By usig the factorisatio for 5 +, the value of 4 si cos comes out to be (A) 4 (B) /4 (C) (D). If ( + ) = ( )( p + )... ( p + ) where N & p, p... p are real umbers the p + p p = (A) (B) 0 (C) ta(/) (D) oe of these Comprehesio # : I the figure = r is circumcircle of ABC.D, & F are the middle ( a ) A poits of the sides BC, CA & AB respectively, AD produced to meet the circle at L. If CAD =, AD = x, BD = y ad altitude of ABC from A meet the circle = r at M, a, b & c are affixes of vertices A, B & C respectively. O the basis of above iformatio, aswer the followig questios :. Area of the ABC is equal to - ( ) B b O P M D L C() c (A) xy cos ( + C) (B) (x + y) si (C) xy si ( + C) (D) xy si ( + C). Affix of M is - (A) b e ib (B) b e i( B) (C) b e ib (D) b e ib. Affix of L is - (A) b e i(a ) (B) b e i(a ) (C) b e i(a ) (D) b ei(a ) MISCLLANOUS TYP QUSTION ANSWR KY X R CI S - Match the Colum. (A) (s), (B) (p), (C) (q), (D) (r). (A) (q), (B) (p), (C) (q, s), (D) (r) Assertio & Reaso. D. A. B 4. C 5. B Comprehesio Based Que st ios Comprehesi o # :. A. C. A Comprehesio # :. C. B. A 54 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

29 J-Mathematics XRCIS - 04 [A] CONCPTUAL SUBJCTIV XRCIS. Fid the modulus, argumet ad the pricipal argumet of the complex umbers. (a) = + cos i si 9 9 (b) (ta i) (c) = 5 i 5 i 5 i 5 i. Give that x, y R, solve : 4x² + xy + (xy x²)i = 4y² (x /) + (xy y²)i. Let ad be two complex umbers such that 4. If i + + i = 0, the prove that =. = ad, fid. 5. If A, B ad C are the agle of a triagle D = e e e ia ic ib e e e ic ib ia e e e ib ia ic where i =, the fid the value of D. NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 6. For complex umbers &, prove that, if ad oly if, = or 7. Let, be complex umbers with = =, prove that Iterpret the followig locii i C. (a) < i < (b) Re i 4 ( i) i (c) Arg ( + i) Arg ( i) = / (d) Arg ( a) = / where a = + 4i. 9. Let A = {a R the equatio ( + i)x ( + i)x + (5 4i)x + a = 0} has at least oe real root. Fid the value of a. aa 0. ABCD is a rhombus i the Argad plae. If the affixes of the vertices be,,, 4 ad take i ati-clockwise sese ad CBA = /, show that (a) = ( + i ) + ( i ) & (b) 4 = ( i ) + ( + i ). P is a poit o the Argad plae. O the circle with OP as diameter two poits Q & R are take such that POQ = QOR =. If 'O' is the origi & P, Q & R are represeted by the complex umbers Z, Z & Z respectively, show that : Z cos = Z. Z cos.. Let A ; B ; C are three complex umbers deotig the vertices of a acuteagled triagle. If the origi O is the orthocetre of the triagle, the prove that.. (a) If is a imagiary cube root of uity the prove that : ( + ) ( + 4 ) ( )... to factors =. (b) If is a complex cube root of uity, fid the value of ; ( + ) ( + ) ( + 4 ) ( + 8 )... to factors. 55

30 J-Mathematics 4. If the biquadratic x 4 + ax + bx + cx + d = 0 (a, b, c, d R) has 4 o real roots, two with sum + 4i ad the other two with product + i. Fid the value of 'b'. 5. If x = + i ; y = i & =, the prove that x p + y p = p for every prime p >. CONCPTUAL SUBJCTIV XRCIS ANSWR KY X R C IS - 4 ( A ). (a) Pricipal Arg = ; = cos ; Arg = k (b) Modulus = sec, Arg = +( ), Pricipal Arg = ( ) (c) Pricipal value of Arg = Pricipal value of Arg = & = & =, Arg =, I, Arg =, I 56 k I. x = K, y = K K R (a) The regio betwee the cocetric circles with cetre at (0, ) & radii & uits (b) regio outside or o the circle with cetre + i ad radius (c) semi circle (i the st & 4th quadrat) x² + y² = (d) a ray emaatig from the poit ( + 4i) directed away from the origi & havig equatio x y (b) oe if is eve ; ² if is odd 4. 5 NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

31 J-Mathematics XRCIS - 04 [B] BRAIN STORMING SUBJCTIV XRCIS. (a) Let = x + iy be a complex umber, where x ad y are real umbers. Let A ad B be the sets defied by. If A = { } ad B = { ( i) + ( + i) 4}. Fid the area of the regio A B. (b) For all real umbers x, let the mappig f(x) = x i, where i =. If there exist real umbers a, b, c ad d for which f(a), f(b), f(c) ad f(d) form a square o the complex plae. Fid the area of the square. p q r q r p 0 ; where p, q, r are the moduli of o-ero complex umbers u, v, w respectively, prove r p q that, arg w v w u = arg v u.. For x (0, /) ad si x =, if si(x) a b b the fid the value of (a + b + c), where a, b, c are c 0 positive itegers. (You may use the fact that si x = 4. If, are the roots of the equatio a + b + c = 0, with a, b, c > 0 ; b > 4ac > b ; third quadrat ; secod quadrat i the argad's plae the, show that e ix e i ix ) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 arg / cos b 4ac 5. If Z r, r =,,,... m, m N are the roots of the equatio Z m + Z m + Z m Z + = 0 the prove that m Z = m r r 6. If ( + x) = C 0 + C x C x ( N), prove that : (a) C 0 + C 4 + C = (c) C + C 6 + C = (e) C 0 + C + C 6 + C = / cos 4 / cos 4 cos (b) C + C 5 + C = (d) C + C 7 + C +... = / si 4 / si 4 7. Prove that : (a) cos x + C cos x + C cos x C cos ( + ) x =. cos x. cos x (b) si x + C si x + C si x C si ( + ) x =. cos x. si x 8. The poits A, B, C depict the complex umbers,, respectively o a complex plae & the agle B & C of the triagle ABC are each equal to ( ). Show that : ( ) 4( )( ) si 9. valuate : q q (p ) si i cos 0. p q 0. Let a, b, c be distict complex umbers such that p a b c b =k. Fid the value of k. c a BRAIN STORMING SUBJCTIV XRCIS ANSWR KY X R C I S - 4 ( B ). (a) (b) / ( i) 0. or 57

32 J-Mathematics XRCIS - 05 [A] J-[MAIN] : PRVIOUS YAR QUSTIONS. The iequality 4 < represets the followig regio [AI - 00 ] () Re() > 0 () Re() < 0 () Re() > (4) oe of these. Let ad are two o-ero complex umbers such that = ad arg + arg =, the equal to [AI - 00 ] () () () (4). Let ad be two roots of the equatio + a + b = 0, beig complex, Further, assume that the origi, ad form a equilateral triagle. the- [AI - 00 ] () a = b () a = b () a = b (4) a = 4b 4. If ad are two o-ero complex umbers such that =, ad Arg() Arg() = /, the is equal to [AI - 00 ] 5. If () () () i (4) i + i i x =, the [AI - 00 ] () x = 4, where is ay positive iteger () x =, where is ay positive iteger () x = 4 +, where is ay positive iteger (4) x = +, where is ay positive iteger 6. Let, w be complex umbers such that + i w = 0 ad arg w =. The arg equals [AI ] () /4 () / () /4 (4) 5/4 7. If = +, the lies o [AI ] () the real axis () the imagiary axis () a circle (4) a ellipse 8. If = x iy ad / = p + iq, the x y + p q is equal to- [AI ] (p + q ) () () () (4) 9. If ad are two o ero complex umbers such that + = + the arg arg is equal to- [AI ] () () () (4) 0 0. If w = ad w = the lies o [AI ] i () a circle () a ellipse () a parabola (4) a straight lie. If + 4, the the maximum value of + is- [AI ] () 4 () 0 () 6 (4) 0. The cojugate of a complex umber is () i () i, the that complex umber is- [AI ] i 4. If Z, the the maximum value of Z is equal to :- [AI ] Z 58 () i (4) i () () + () + (4) 5 + NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

33 J-Mathematics 4. The umber of complex umbers such that = + = i equals :- [AI - 0 0] () 0 () () (4) 5. Let, be real ad be a complex umber. If + + = 0 has two distict roots o the lie Re =, the it is ecessary that :- [AI - 0 ] () () (, ) () (0,) (4) (,0) 6. If () is a cube root of uity, ad ( +) 7 = A + B. The (A, B) equals :- [AI - 0 ] () (, 0) () (, ) () (0, ) (4) (, ) 7. If ad is real, the the poit represeted by the complex umber lies : [AI - 0 ] () o the imagiary axis. () either o the real axis or o a circle passig through the origi. () o a circle with cetre at the origi. (4) either o the real axis or o a circle ot passig through the origi. 8. If is a complex umber of uit modulus ad argumet, the arg equals [J (Mai)-0] () () () (4) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 PRVIOUS YARS QUSTIONS ANSWR KY XRCIS-5 [A] Q u e A s Q u e A s 4 59

34 J-Mathematics XRCIS - 05 [B] J-[ADVANCD] : PRVIOUS YAR QUSTIONS. (a) If,, are complex umbers such that the + + is - 60 (A) equal to (B) less tha (C) greater tha (D) equal to (b) If arg () < 0, the arg ( ) arg () = [J 000 Screeig) +M out of 5] (A) (B) (C). (a) The complex umbers, ad satisfyig i (D) are the vertices of a triagle which is - (A) of area ero (B) right-agled isosceles (C) equilateral (D) obtuse-agled isosceles (b) Let ad be th roots of uity which subted a right agle at the origi. The must be of the form. (a) Let (A) 4k + (B) 4k + (C) 4k + (D) 4k i. The the value of the determiat [J 00 (Screeig) +M out of 5] 4 is - [J 0 (Screeig) M] (A) (B) ( ) (C) (D) ( ) ( b ) For all complex umbers, satisfyig = ad 4i = 5, the miimum value of is [J 0 (Screeig) M] (A) 0 (B) (C) 7 (D) 7 ( c ) Let a complex umber,, be a root of the equatio p+q p q + =0 where p,q are distict primes. Show that either p - = 0 or q- =0, but ot both together. [J 0 (Mais) 5M] 4. If = ad (where ), the Re (w) equals [J 0 (Screeig) M] (A) 0 (B) (C). (D) 5. If ad are two complex umbers such that < ad > the show that [J 0 (mais) M out of 60)] r 6. Show that there exists o complex umber such that ad a r r where a i < for i =,,... [J 0 (mais) M out of 60)] 7. The least positive value of for which ( + ) = ( + 4 ), where is a o real cube root of uity is - (A) (B) (C) 6 (D) 4 [J 04 (screeig) M] 8. Fid the cetre ad radius formed by all the poits represeted by = x + i y satisfyig the relatio K (K ) where & are costat complex umbers, give by i & i [J 04 (Mais) ( out of 60)] 9. If a, b, c are itegers ot all equal ad is cube root of uity ( ) the the miimum value of a + b + c is - [J 05 (screeig) M] (A) 0 (B) (C) (D) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

35 J-Mathematics 0. Area of shaded regio belogs to - [J 05 (screeig) M] (A) : + >, arg ( + ) < /4 (B) : >, arg ( ) < /4 A (,0) P(, ) /4 (,0) (C) : + <, arg ( + ) < / (D) : <, arg ( ) < / Q(, ) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65. If oe of the vertices of the square circumscribig the circle = is i. Fid the other vertices of square. [J 05 (Mais) 4 out of 60]. If w = i where 0 ad, satisfies the coditio that w w is purely real, the the set of values of is - [J 06, M] (A) { : =} (B) { : = } (C) { : } (D) { : =, }. A ma walks a distace of uits from the origi towards the orth-east (N 45 ) directio. From there, he walks a distace of 4 uits towards the orth-west (N 45 W) directio to reach a poit P. The the positio of P i the Argad plae is : [J 07, M] (A) e i/4 + 4i (B) ( 4i)e i/4 (C) (4 + i)e i/4 (D) ( + 4i)e i/4 4. If = ad ±, the all the values of lie o : [J 07, M] (A) a lie ot passig through the origi (B) = (C) the x-axis (D) the y-axis Comprehesi o (for 5 to 7) : Let A, B, C be three sets of complex umbers as defied below A : Im B : i C : Re(( i)) [J 008, 4M, M] 5. The umber of elemets i the set A B C is - (A) 0 (B) (C) (D) 6. Let be ay poit i A B C. The + i + 5 i lies betwee - (A) 5 ad 9 (B) 0 ad 4 (C) 5 ad 9 (D) 40 ad Let be ay poit i A B C ad let be ay poit satisfyig i <. The, + lies betwee - (A) 6 ad (B) ad 6 (C) 6 ad 6 (D) ad 9 8. A particle P starts from the poit 0 = + i, where i =. It moves first horiotally away from origi by 5 uits ad the vertically away from origi by uits to reach a poit. From the particle moves uits i the directio of the vector ˆi ˆj ad the it moves through a agle i aticlockwise directio o a circle with cetre at origi, to reach a poit. The poit is give by - [J 008, M, M] (A) 6 + 7i (B) 7 + 6i (C) 7 + 6i (D) 6 + 7i 9. Let = cos + i si. The the value of 5 m Im( ) at = is - [J 009, M, M] m (A) (B) (C) (D) si si si 4 si 0. Let = x + iy be a complex umber where x ad y are itegers. The the area of the rectagle whose vertices are the roots of the equatio 50 is - [J 009, M, M] (A) 48 (B) (C) 40 (D) 80 6

36 J-Mathematics. Match the coics i Colum I with the statemets/ expressios i Colum II. [J 009, 8M] Colum I 6 Colum II (A) Circle (P) The locus of the poit (h, k) for which the lie (B) Parabola hx + ky = touches the circle x + y = 4 (C) llipse (Q) Poits i the complex plae satisfyig + = ± (D) Hyperbola (R) Poits of the coic have parametric represetatio (S) t x t, y = t t The eccetricity of the coic lies i the iterval x < (T) Poits i the complex plae satisfyig Re ( + ) = +. Let ad be two distict complex umbers ad let = ( t) + t for some real umber t with 0 < t <. If Arg(w) deotes the pricipal argumet of a oero complex umber w, the (A) + = (B) Arg( ) = Arg( ) [J 0, M] (C) 0. Let be the complex umber 0 (D) Arg( ) = Arg( ) cos i si. The the umber of distict complex umbers satisfyig is equal to [J 0, M] 4. Match the statemets i Colum-I with those i Colum-II. [J 0, 8M] [Note : Here takes values i the complex plae ad Im ad Re deote, respectively, the imagiary part ad the real part of.] Colum I Colum II (A) The set of poits satisfyig i i (p) a ellipse with eccetricity 4 5 is cotaied i or equal to (B) The set of poits satisfyig (q) the set of poits satisfyig Im = = 0 is cotaied i or equal to (C) If w =, the the set of poits (t) the set of poits satisfyig Im < w is cotaied i or equal to w (D) If w =, the the set of poits (s) the set of poits satisfyig Re w is cotaied i or equal to (t) the set of poits satisfyig w 5. Comprehesio ( questios together) Let a,b ad c be three real umbers satisfyig 9 7 a b c ( i ) If the poit P(a,b,c), with referece to (), lies o the plae x + y + =, the the value of 7a+b+c is (A) 0 (B) (C) 7 (D) 6...() NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65

37 J-Mathematics (ii) Let be a solutio of x = 0 with Im() > 0. If a = with b ad c satisfyig (), (iii) the the value of a b c is equal to - (A) (B) (C) (D) Let b = 6, with a ad c satisfyig (). If ad are the roots of the quadratic equatio ax + bx + c = 0, the 0 is - (A) 6 (B) 7 (C) 6 7 (D) [J 0, ++] 6. If is ay complex umber satisfyig i <, the the miimum value of 6 + 5i 7. Let is [J 0, 4M] e i /, ad a, b, c, x, y, be o-ero complex umbers such that a + b + c = x a + b + c = y a + b + c =. x y The the value of is [J 0, 4M] a b c 8. Match the statemets give i Colum I with the values give i Colum II Colum I Colum II (A) If a ˆj k, ˆ b ˆj k ˆ ad c k ˆ form a triagle, (p) 6 the the iteral agle of the triagle betwee a ad b is b a (B) If (ƒ(x) x)dx a b, the the value of (C) The value of / 6 sec( x)dx is ƒ is (q) 6 (r) NOD6\\Data\04\Kota\J-Advaced\SMP\Maths\Uit#\g\0.Complex umber.p65 (D) The maximum value of =, is give by Arg for (s) (t) [J 0, +++M] 9. Match the statemets give i Colum I with the itervals/uio of itervals give i Colum II Colum I Colum II (A) The set i Re : is a complex umber, =, is (B) (C) The domai of the fuctio If ta ƒ(x) 8() x si ( x ) ƒ( ) ta ta, the the set ta (p) (, ) (, ) is (q) (,0) (0, ) ƒ( ) : 0 is (r) [, ) (D) If ƒ(x) = x / (x 0), x 0, the ƒ(x) is icreasig i (s) (, ] [, ) (t) (,0] [, ) [J 0, +++M] 6

EXERCISE - 01 CHECK YOUR GRASP

EXERCISE - 01 CHECK YOUR GRASP J-Mathematics XRCIS - 0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). The maximum value of the sum of the A.P. 0, 8, 6,,... is - 68 60 6. Let T r be the r th term of a A.P. for r =,,,...

More information

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018)

JEE(Advanced) 2018 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 20 th MAY, 2018) JEE(Advaced) 08 TEST PAPER WITH SOLUTION (HELD ON SUNDAY 0 th MAY, 08) PART- : JEE(Advaced) 08/Paper- SECTION. For ay positive iteger, defie ƒ : (0, ) as ƒ () j ta j j for all (0, ). (Here, the iverse

More information

Objective Mathematics

Objective Mathematics . If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic

More information

COMPLEX NUMBERS AND DE MOIVRE'S THEOREM SYNOPSIS. Ay umber of the form x+iy where x, y R ad i = - is called a complex umber.. I the complex umber x+iy, x is called the real part ad y is called the imagiary

More information

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms. [ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural

More information

Complex Numbers Solutions

Complex Numbers Solutions Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i

More information

APPENDIX F Complex Numbers

APPENDIX F Complex Numbers APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios

More information

Objective Mathematics

Objective Mathematics 6. If si () + cos () =, the is equal to :. If <

More information

Poornima University, For any query, contact us at: ,18

Poornima University, For any query, contact us at: ,18 AIEEE/1/MAHS 1 S. No Questios Solutios Q.1 he circle passig through (1, ) ad touchig the axis of x at (, ) also passes through the poit (a) (, ) (b) (, ) (c) (, ) (d) (, ) Q. ABCD is a trapezium such that

More information

Presentation of complex number in Cartesian and polar coordinate system

Presentation of complex number in Cartesian and polar coordinate system a + bi, aεr, bεr i = z = a + bi a = Re(z), b = Im(z) give z = a + bi & w = c + di, a + bi = c + di a = c & b = d The complex cojugate of z = a + bi is z = a bi The sum of complex cojugates is real: z +

More information

SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2

SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2 Class-Jr.X_E-E SIMPLE HOLIDAY PACKAGE CLASS-IX MATHEMATICS SUB BATCH : E-E SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY. siθ+cosθ + siθ cosθ = ) ) ). If a cos q, y bsi q, the a y b ) ) ). The value

More information

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=

More information

Solutions for May. 3 x + 7 = 4 x x +

Solutions for May. 3 x + 7 = 4 x x + Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits

More information

WBJEE Answer Keys by Aakash Institute, Kolkata Centre

WBJEE Answer Keys by Aakash Institute, Kolkata Centre WBJEE - 7 Aswer Keys by, Kolkata Cetre MATHEMATICS Q.No. B A C B A C A B 3 D C B B 4 B C D D 5 D A B B 6 C D B B 7 B C C A 8 B B A A 9 A * B D C C B B D A A D B B C B 3 A D D D 4 C B A A 5 C B B B 6 C

More information

Appendix F: Complex Numbers

Appendix F: Complex Numbers Appedix F Complex Numbers F1 Appedix F: Complex Numbers Use the imagiary uit i to write complex umbers, ad to add, subtract, ad multiply complex umbers. Fid complex solutios of quadratic equatios. Write

More information

02 - COMPLEX NUMBERS Page 1 ( Answers at the end of all questions ) l w l = 1, then z lies on

02 - COMPLEX NUMBERS Page 1 ( Answers at the end of all questions ) l w l = 1, then z lies on 0 - COMPLEX NUMBERS Page ( ) If the cube roots of uity are,,, the the roots of the equatio ( x - ) + 8 = 0 are ( a ) -, - +, - - ( b ) -, -, -, ( c ) -, -, - ( d ) -, +, + [ AIEEE 005 ] ( ) If z ad z are

More information

2 Geometric interpretation of complex numbers

2 Geometric interpretation of complex numbers 2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that

More information

REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.

REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains. The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values of the variable it cotais The relatioships betwee

More information

Complex Numbers. Brief Notes. z = a + bi

Complex Numbers. Brief Notes. z = a + bi Defiitios Complex Numbers Brief Notes A complex umber z is a expressio of the form: z = a + bi where a ad b are real umbers ad i is thought of as 1. We call a the real part of z, writte Re(z), ad b the

More information

SEQUENCE AND SERIES NCERT

SEQUENCE AND SERIES NCERT 9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of

More information

BITSAT MATHEMATICS PAPER III. For the followig liear programmig problem : miimize z = + y subject to the costraits + y, + y 8, y, 0, the solutio is (0, ) ad (, ) (0, ) ad ( /, ) (0, ) ad (, ) (d) (0, )

More information

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1

Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1 Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate

More information

2) 3 π. EAMCET Maths Practice Questions Examples with hints and short cuts from few important chapters

2) 3 π. EAMCET Maths Practice Questions Examples with hints and short cuts from few important chapters EAMCET Maths Practice Questios Examples with hits ad short cuts from few importat chapters. If the vectors pi j + 5k, i qj + 5k are colliear the (p,q) ) 0 ) 3) 4) Hit : p 5 p, q q 5.If the vectors i j

More information

REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.

REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains. the Further Mathematics etwork wwwfmetworkorguk V 07 The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values

More information

+ {JEE Advace 03} Sept 0 Name: Batch (Day) Phoe No. IT IS NOT ENOUGH TO HAVE A GOOD MIND, THE MAIN THING IS TO USE IT WELL Marks: 00. If A (α, β) = (a) A( α, β) = A( α, β) (c) Adj (A ( α, β)) = Sol : We

More information

TEACHER CERTIFICATION STUDY GUIDE

TEACHER CERTIFICATION STUDY GUIDE COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra

More information

Complex Analysis Spring 2001 Homework I Solution

Complex Analysis Spring 2001 Homework I Solution Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle

More information

Q.11 If S be the sum, P the product & R the sum of the reciprocals of a GP, find the value of

Q.11 If S be the sum, P the product & R the sum of the reciprocals of a GP, find the value of Brai Teasures Progressio ad Series By Abhijit kumar Jha EXERCISE I Q If the 0th term of a HP is & st term of the same HP is 0, the fid the 0 th term Q ( ) Show that l (4 36 08 up to terms) = l + l 3 Q3

More information

MOCK TEST - 02 COMMON ENTRANCE TEST 2012 SUBJECT: MATHEMATICS Time: 1.10Hrs Max. Marks 60 Questions 60. then x 2 =

MOCK TEST - 02 COMMON ENTRANCE TEST 2012 SUBJECT: MATHEMATICS Time: 1.10Hrs Max. Marks 60 Questions 60. then x 2 = MOCK TEST - 0 COMMON ENTRANCE TEST 0 SUBJECT: MATHEMATICS Time:.0Hrs Max. Marks 60 Questios 60. The value of si cot si 3 cos sec + + 4 4 a) 0 b) c) 4 6 + x x. If Ta - α + x + x the x a) cos α b) Taα c)

More information

De Moivre s Theorem - ALL

De Moivre s Theorem - ALL De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.

More information

STUDY PACKAGE. Subject : Mathematics Topic : The Point & Straight Lines

STUDY PACKAGE. Subject : Mathematics Topic : The Point & Straight Lines fo/u fopkjr Hkh# tu] ugha vkjehks dke] foifr s[k NksM+s rqjar e/;e eu dj ';kea iq#"k flag ladyi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA jfpr% ekuo /kez izksrk l~xq# Jh jknksm+klth

More information

MTH Assignment 1 : Real Numbers, Sequences

MTH Assignment 1 : Real Numbers, Sequences MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a

More information

WBJEE MATHEMATICS

WBJEE MATHEMATICS WBJEE - 06 MATHEMATICS Q.No. 0 A C B B 0 B B A B 0 C A C C 0 A B C C 05 A A B C 06 B C B C 07 B C A D 08 C C C A 09 D D C C 0 A C A B B C B A A C A B D A A A B B D C 5 B C C C 6 C A B B 7 C A A B 8 C B

More information

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j

The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.

More information

Mathematics Extension 2 SOLUTIONS

Mathematics Extension 2 SOLUTIONS 3 HSC Examiatio Mathematics Extesio SOLUIONS Writte by Carrotstics. Multiple Choice. B 6. D. A 7. C 3. D 8. C 4. A 9. B 5. B. A Brief Explaatios Questio Questio Basic itegral. Maipulate ad calculate as

More information

( ) D) E) NOTA

( ) D) E) NOTA 016 MAΘ Natioal Covetio 1. Which Greek mathematicia do most historias credit with the discovery of coic sectios as a solutio to solvig the Delia problem, also kow as doublig the cube? Eratosthees Meaechmus

More information

Lesson 10: Limits and Continuity

Lesson 10: Limits and Continuity www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals

More information

BRAIN TEASURES TRIGONOMETRICAL RATIOS BY ABHIJIT KUMAR JHA EXERCISE I. or tan &, lie between 0 &, then find the value of tan 2.

BRAIN TEASURES TRIGONOMETRICAL RATIOS BY ABHIJIT KUMAR JHA EXERCISE I. or tan &, lie between 0 &, then find the value of tan 2. EXERCISE I Q Prove that cos² + cos² (+ ) cos cos cos (+ ) ² Q Prove that cos ² + cos (+ ) + cos (+ ) Q Prove that, ta + ta + ta + cot cot Q Prove that : (a) ta 0 ta 0 ta 60 ta 0 (b) ta 9 ta 7 ta 6 + ta

More information

Solving equations (incl. radical equations) involving these skills, but ultimately solvable by factoring/quadratic formula (no complex roots)

Solving equations (incl. radical equations) involving these skills, but ultimately solvable by factoring/quadratic formula (no complex roots) Evet A: Fuctios ad Algebraic Maipulatio Factorig Square of a sum: ( a + b) = a + ab + b Square of a differece: ( a b) = a ab + b Differece of squares: a b = ( a b )(a + b ) Differece of cubes: a 3 b 3

More information

AIEEE 2004 (MATHEMATICS)

AIEEE 2004 (MATHEMATICS) AIEEE 00 (MATHEMATICS) Importat Istructios: i) The test is of hours duratio. ii) The test cosists of 75 questios. iii) The maimum marks are 5. iv) For each correct aswer you will get marks ad for a wrog

More information

Stanford Math Circle January 21, Complex Numbers

Stanford Math Circle January 21, Complex Numbers Staford Math Circle Jauary, 007 Some History Tatiaa Shubi (shubi@mathsjsuedu) Complex Numbers Let us try to solve the equatio x = 5x + x = is a obvious solutio Also, x 5x = ( x )( x + x + ) = 0 yields

More information

3sin A 1 2sin B. 3π x is a solution. 1. If A and B are acute positive angles satisfying the equation 3sin A 2sin B 1 and 3sin 2A 2sin 2B 0, then A 2B

3sin A 1 2sin B. 3π x is a solution. 1. If A and B are acute positive angles satisfying the equation 3sin A 2sin B 1 and 3sin 2A 2sin 2B 0, then A 2B 1. If A ad B are acute positive agles satisfyig the equatio 3si A si B 1 ad 3si A si B 0, the A B (a) (b) (c) (d) 6. 3 si A + si B = 1 3si A 1 si B 3 si A = cosb Also 3 si A si B = 0 si B = 3 si A Now,

More information

Mathematics Extension 2

Mathematics Extension 2 009 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Etesio Geeral Istructios Readig time 5 miutes Workig time hours Write usig black or blue pe Board-approved calculators may be used A table of stadard

More information

JEE ADVANCED 2013 PAPER 1 MATHEMATICS

JEE ADVANCED 2013 PAPER 1 MATHEMATICS Oly Oe Optio Correct Type JEE ADVANCED 0 PAPER MATHEMATICS This sectio cotais TEN questios. Each has FOUR optios (A), (B), (C) ad (D) out of which ONLY ONE is correct.. The value of (A) 5 (C) 4 cot cot

More information

First selection test, May 1 st, 2008

First selection test, May 1 st, 2008 First selectio test, May st, 2008 Problem. Let p be a prime umber, p 3, ad let a, b be iteger umbers so that p a + b ad p 2 a 3 + b 3. Show that p 2 a + b or p 3 a 3 + b 3. Problem 2. Prove that for ay

More information

Patterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry

Patterns in Complex Numbers An analytical paper on the roots of a complex numbers and its geometry IB MATHS HL POTFOLIO TYPE Patters i Complex Numbers A aalytical paper o the roots of a complex umbers ad its geometry i Syed Tousif Ahmed Cadidate Sessio Number: 0066-009 School Code: 0066 Sessio: May

More information

Chapter 1. Complex Numbers. Dr. Pulak Sahoo

Chapter 1. Complex Numbers. Dr. Pulak Sahoo Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia E-mail : sahoopulak1@gmail.com 1 Module-2: Stereographic Projectio 1 Euler

More information

ANSWERSHEET (TOPIC = ALGEBRA) COLLECTION #2

ANSWERSHEET (TOPIC = ALGEBRA) COLLECTION #2 Teko Classes IITJEE/AIEEE Maths by SUHAAG SIR, Bhopal, Ph (0755) 00 000 www.tekoclasses.com ANSWERSHEET (TOPIC ALGEBRA) COLLECTION # Questio Type A.Sigle Correct Type Q. (B) Sol ( 5 7 ) ( 5 7 9 )!!!! C

More information

PAPER : IIT-JAM 2010

PAPER : IIT-JAM 2010 MATHEMATICS-MA (CODE A) Q.-Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure

More information

AH Checklist (Unit 3) AH Checklist (Unit 3) Matrices

AH Checklist (Unit 3) AH Checklist (Unit 3) Matrices AH Checklist (Uit 3) AH Checklist (Uit 3) Matrices Skill Achieved? Kow that a matrix is a rectagular array of umbers (aka etries or elemets) i paretheses, each etry beig i a particular row ad colum Kow

More information

Unit 4: Polynomial and Rational Functions

Unit 4: Polynomial and Rational Functions 48 Uit 4: Polyomial ad Ratioal Fuctios Polyomial Fuctios A polyomial fuctio y px ( ) is a fuctio of the form p( x) ax + a x + a x +... + ax + ax+ a 1 1 1 0 where a, a 1,..., a, a1, a0are real costats ad

More information

ANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A.

ANSWERS SOLUTIONS iiii i. and 1. Thus, we have. i i i. i, A. 013 ΜΑΘ Natioal Covetio ANSWERS (1) C A A A B (6) B D D A B (11) C D D A A (16) D B A A C (1) D B C B C (6) D C B C C 1. We have SOLUTIONS 1 3 11 61 iiii 131161 i 013 013, C.. The powers of i cycle betwee

More information

Assignment ( ) Class-XI. = iii. v. A B= A B '

Assignment ( ) Class-XI. = iii. v. A B= A B ' Assigmet (8-9) Class-XI. Proe that: ( A B)' = A' B ' i A ( BAC) = ( A B) ( A C) ii A ( B C) = ( A B) ( A C) iv. A B= A B= φ v. A B= A B ' v A B B ' A'. A relatio R is dified o the set z of itegers as:

More information

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =

Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t = Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,

More information

Mathematics Extension 2

Mathematics Extension 2 004 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Etesio Geeral Istructios Readig time 5 miutes Workig time hours Write usig black or blue pe Board-approved calculators may be used A table of stadard

More information

SEQUENCE AND SERIES. Contents. Theory Exercise Exercise Exercise Exercise

SEQUENCE AND SERIES. Contents. Theory Exercise Exercise Exercise Exercise SEQUENCE AND SERIES Cotets Topic Page No. Theory 0-0 Exercise - 05-09 Exercise - 0-3 Exercise - 3-7 Exercise - 8-9 Aswer Key 0 - Syllabus Arithmetic, geometric ad harmoic progressios, arithmetic, geometric

More information

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016

18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016 18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam

More information

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get

Substitute these values into the first equation to get ( z + 6) + ( z + 3) + z = 27. Then solve to get Problem ) The sum of three umbers is 7. The largest mius the smallest is 6. The secod largest mius the smallest is. What are the three umbers? [Problem submitted by Vi Lee, LCC Professor of Mathematics.

More information

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials

3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered

More information

Regn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: ,

Regn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: , . Sectio-A cotais 30 Multiple Choice Questios (MCQ). Each questio has 4 choices (a), (b), (c) ad (d), for its aswer, out of which ONLY ONE is correct. From Q. to Q.0 carries Marks ad Q. to Q.30 carries

More information

September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1

September 2012 C1 Note. C1 Notes (Edexcel) Copyright   - For AS, A2 notes and IGCSE / GCSE worksheets 1 September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright

More information

NBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian?

NBHM QUESTION 2007 Section 1 : Algebra Q1. Let G be a group of order n. Which of the following conditions imply that G is abelian? NBHM QUESTION 7 NBHM QUESTION 7 NBHM QUESTION 7 Sectio : Algebra Q Let G be a group of order Which of the followig coditios imply that G is abelia? 5 36 Q Which of the followig subgroups are ecesarily

More information

We will conclude the chapter with the study a few methods and techniques which are useful

We will conclude the chapter with the study a few methods and techniques which are useful Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs

More information

Coffee Hour Problems of the Week (solutions)

Coffee Hour Problems of the Week (solutions) Coffee Hour Problems of the Week (solutios) Edited by Matthew McMulle Otterbei Uiversity Fall 0 Week. Proposed by Matthew McMulle. A regular hexago with area 3 is iscribed i a circle. Fid the area of a

More information

C. Complex Numbers. x 6x + 2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions

C. Complex Numbers. x 6x + 2 = 0. This equation was known to have three real roots, given by simple combinations of the expressions C. Complex Numbers. Complex arithmetic. Most people thik that complex umbers arose from attempts to solve quadratic equatios, but actually it was i coectio with cubic equatios they first appeared. Everyoe

More information

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?

More information

R is a scalar defined as follows:

R is a scalar defined as follows: Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad

More information

1. By using truth tables prove that, for all statements P and Q, the statement

1. By using truth tables prove that, for all statements P and Q, the statement Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3

More information

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007 UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =

More information

Chapter 7: The z-transform. Chih-Wei Liu

Chapter 7: The z-transform. Chih-Wei Liu Chapter 7: The -Trasform Chih-Wei Liu Outlie Itroductio The -Trasform Properties of the Regio of Covergece Properties of the -Trasform Iversio of the -Trasform The Trasfer Fuctio Causality ad Stability

More information

VITEEE 2018 MATHEMATICS QUESTION BANK

VITEEE 2018 MATHEMATICS QUESTION BANK VITEEE 8 MTHEMTICS QUESTION BNK, C = {,, 6}, the (B C) Ques. Give the sets {,,},B {, } is {} {,,, } {,,, } {,,,,, 6} Ques. s. d ( si cos ) c ta log( ta 6 Ques. The greatest umer amog 9,, 7 is ) c c cot

More information

Analytic Continuation

Analytic Continuation Aalytic Cotiuatio The stadard example of this is give by Example Let h (z) = 1 + z + z 2 + z 3 +... kow to coverge oly for z < 1. I fact h (z) = 1/ (1 z) for such z. Yet H (z) = 1/ (1 z) is defied for

More information

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of

More information

INEQUALITIES BJORN POONEN

INEQUALITIES BJORN POONEN INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad

More information

6.003 Homework #3 Solutions

6.003 Homework #3 Solutions 6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the

More information

4755 Mark Scheme June Question Answer Marks Guidance M1* Attempt to find M or 108M -1 M 108 M1 A1 [6] M1 A1

4755 Mark Scheme June Question Answer Marks Guidance M1* Attempt to find M or 108M -1 M 108 M1 A1 [6] M1 A1 4755 Mark Scheme Jue 05 * Attempt to fid M or 08M - M 08 8 4 * Divide by their determiat,, at some stage Correct determiat, (A0 for det M= 08 stated, all other OR 08 8 4 5 8 7 5 x, y,oe 8 7 4xy 8xy dep*

More information

Solutions. tan 2 θ(tan 2 θ + 1) = cot6 θ,

Solutions. tan 2 θ(tan 2 θ + 1) = cot6 θ, Solutios 99. Let A ad B be two poits o a parabola with vertex V such that V A is perpedicular to V B ad θ is the agle betwee the chord V A ad the axis of the parabola. Prove that V A V B cot3 θ. Commet.

More information

Mathematics Extension 1

Mathematics Extension 1 016 Bored of Studies Trial Eamiatios Mathematics Etesio 1 3 rd ctober 016 Geeral Istructios Total Marks 70 Readig time 5 miutes Workig time hours Write usig black or blue pe Black pe is preferred Board-approved

More information

Fundamental Concepts: Surfaces and Curves

Fundamental Concepts: Surfaces and Curves UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat

More information

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense,

3. Z Transform. Recall that the Fourier transform (FT) of a DT signal xn [ ] is ( ) [ ] = In order for the FT to exist in the finite magnitude sense, 3. Z Trasform Referece: Etire Chapter 3 of text. Recall that the Fourier trasform (FT) of a DT sigal x [ ] is ω ( ) [ ] X e = j jω k = xe I order for the FT to exist i the fiite magitude sese, S = x [

More information

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n

We are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at

More information

PUTNAM TRAINING INEQUALITIES

PUTNAM TRAINING INEQUALITIES PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca

More information

IYGB. Special Extension Paper E. Time: 3 hours 30 minutes. Created by T. Madas. Created by T. Madas

IYGB. Special Extension Paper E. Time: 3 hours 30 minutes. Created by T. Madas. Created by T. Madas YGB Special Extesio Paper E Time: 3 hours 30 miutes Cadidates may NOT use ay calculator. formatio for Cadidates This practice paper follows the Advaced Level Mathematics Core ad the Advaced Level Further

More information

Complex Numbers Summary

Complex Numbers Summary Complex Numbers Summary What does a complex umber mea? Academic Skills Advice A complex umber has a real part ad a imagiary part (the imagiary part ivolves the square root of a egative umber). We use Z

More information

Objective Mathematics

Objective Mathematics -0 {Mais & Advace} B.E.(CIVIL), MNIT,JAIPUR(Rajastha) Copyright L.K.Sharma 0. Er. L.K.Sharma a egieerig graduate from NIT, Jaipur (Rajastha), {Gold medalist, Uiversity of Rajastha} is a well kow ame amog

More information

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS

SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS . If their mea positios coicide with each other, maimum separatio will be A. Now from phasor diagram, we ca clearly see the phase differece. SAFE HANDS & IIT-ia's PACE ad Aswer : Optio (4) 5. Aswer : Optio

More information

U8L1: Sec Equations of Lines in R 2

U8L1: Sec Equations of Lines in R 2 MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie

More information

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify

RADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL

More information

Math 122 Test 3 - Review 1

Math 122 Test 3 - Review 1 I. Sequeces ad Series Math Test 3 - Review A) Sequeces Fid the limit of the followig sequeces:. a = +. a = l 3. a = π 4 4. a = ta( ) 5. a = + 6. a = + 3 B) Geometric ad Telescopig Series For the followig

More information

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x

More information

MTH112 Trigonometry 2 2 2, 2. 5π 6. cscθ = 1 sinθ = r y. secθ = 1 cosθ = r x. cotθ = 1 tanθ = cosθ. central angle time. = θ t.

MTH112 Trigonometry 2 2 2, 2. 5π 6. cscθ = 1 sinθ = r y. secθ = 1 cosθ = r x. cotθ = 1 tanθ = cosθ. central angle time. = θ t. MTH Trigoometry,, 5, 50 5 0 y 90 0, 5 0,, 80 0 0 0 (, 0) x, 7, 0 5 5 0, 00 5 5 0 7,,, Defiitios: siθ = opp. hyp. = y r cosθ = adj. hyp. = x r taθ = opp. adj. = siθ cosθ = y x cscθ = siθ = r y secθ = cosθ

More information

MODEL TEST PAPER II Time : hours Maximum Marks : 00 Geeral Istructios : (i) (iii) (iv) All questios are compulsory. The questio paper cosists of 9 questios divided ito three Sectios A, B ad C. Sectio A

More information

Review Problems Math 122 Midterm Exam Midterm covers App. G, B, H1, H2, Sec , 8.9,

Review Problems Math 122 Midterm Exam Midterm covers App. G, B, H1, H2, Sec , 8.9, Review Problems Math Midterm Exam Midterm covers App. G, B, H, H, Sec 8. - 8.7, 8.9, 9.-9.7 Review the Cocept Check problems: Page 6/ -, Page 690/- 0 PART I: True-False Problems Ch. 8. Page 6 True-False

More information

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3

(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3 MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special

More information

Addition: Property Name Property Description Examples. a+b = b+a. a+(b+c) = (a+b)+c

Addition: Property Name Property Description Examples. a+b = b+a. a+(b+c) = (a+b)+c Notes for March 31 Fields: A field is a set of umbers with two (biary) operatios (usually called additio [+] ad multiplicatio [ ]) such that the followig properties hold: Additio: Name Descriptio Commutativity

More information

SEQUENCES AND SERIES

SEQUENCES AND SERIES Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces

More information

Chapter 6 Infinite Series

Chapter 6 Infinite Series Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat

More information

GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.5

GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.5 GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 5 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5

More information

Complex Number Theory without Imaginary Number (i)

Complex Number Theory without Imaginary Number (i) Ope Access Library Joural Complex Number Theory without Imagiary Number (i Deepak Bhalchadra Gode Directorate of Cesus Operatios, Mumbai, Idia Email: deepakm_4@rediffmail.com Received 6 July 04; revised

More information

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.

11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4. 11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although

More information