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