Q. If + 0 then which of the following must be true on the complex plane? (A) Re() < 0 (B*) Re() 0 (C) Im() 0 (D) [Hint: ( + ) 0 0 or i 0 or ± i Re() 0] Q. There is only one way to choose real numbers M and N such that when the polynomial 5x + x + x + Mx + N is divided by the polynomial x +, the remainder is 0. If M and N assume these unique values, then M N is (A) 6 (B) (C*) 6 (D) Let P(x) 5x + x + x + Mx + N let R(x) x + if the quotient is Q then P(x) Q(x + ) if x i then P(i) 0 if x i then P( i) 0 hence 5 i + Mi + N 0 hence N + Mi + i N ; M M N 6 Ans. ] Q. The complex number satisfying + + 7i then the value of equals (A*) 65 (B) 69 (C) 9 (D) 5 x + iy x + iy + x y + 7i x + x y... () and y 7... () x + x 9 x + 9 + x x x 8 x x + y 65 Ans.] Q. If and w are satisfying i w 0 and Amp(w) then Amp() is equal to (A) (B) (C) Given i w 0 i w or i w amp() amp(w) amp i...() (D*) also amp(w) amp() + amp(w)...() () + (), gives amp() amp() ; Also amp(w) ] Q.5 If ( + 7i) (p + iq) where p, q I {0}, is purely imaginary then minimum value of is (A) 0 (B) 58 (C) 6 (D*) 6 [Hint : (p 7q) + i(q + 7p) for purely imaginary p 7q p 7 or q (for least value) + 7i p + iq 58(p + q ) 58[7 + 9] 58 (D) ] x y Q.6 If + i (where x, y R, i ) then i i (A) x & y 8 (B*) x & y 8 (C) x & y 6 (D) x & y 8
Q.7 Let 9 + bi where b is non ero real and i. If the imaginary part of and are equal, then b equals (A) 6 (B*) 5 (C) 5 (D) 6 8 b + 8bi 79 + bi 7b b i hence b b 8b and b 8 b 5 Ans. ] Q.8 If is a complex number satisfying the equation ( + i) and, then the locus traced by '' in the complex plane is (A*) x y 0 (B) x + y 0 (C) x y + 0 (D) x + y + 0 We have ( + i) (x ) + (y ) (Put x + iy) x + y (x + y)...() Let h + ik x i y (x iy), so x y x h x y y, k x y (x y) h k x y (from equation ()) Locus of the point (h, k) will be x y Ans. (A) ] Q.9 The digram shows several numbers in the complex plane. The circle is the unit circle centered at the origin. One of these numbers is the reciprocal of F, which is (A) A (B) B (C*) C (D) D Let F as a + bi, a, b R where we see from the diagram that a, b > 0 and a + b > (as F lies outside the unit circle) a bi a b Since i a bi a b a b a, b (real part + ve and imaginary part ve and both less than unity) we see that the reciprocal of F is in IV quadrant, since the real part is positive and the imaginary part is negative. Also, the magnitude of the reciprocal is a ( b) a b a b Thus the only possibility is point C. ] Q.0 If x + iy & i then implies that, in the complex plane i (A) lies on the imaginary axis (B*) lies on the real axis (C) lies on the unit circle (D) none <
i ( i) i w ; w i i + i i lies on the perpendicular bisector of the segment joining (0, ) and (0, ) which is x-axis lies on x-axis Im () is real ] Q. On the complex plane locus of a point satisfying the inequality < denotes (A) region between the concentric circles of radii and centered at (, 0) (B) region between the concentric circles of radii and centered at (, 0) excluding the inner and outer boundaries. (C) region between the concentric circles of radii and centered at (, 0) including the inner and outer boundaries. (D*) region between the concentric circles of radii and centered at (, 0) including the inner boundary and excluding the outer boundary. Q. Let Z (8 + i)sin + (7 + i)cos and Z ( + 8i)sin + ( + 7i)cos are two complex numbers. If Z Z a + ib where a, b R then the largest value of (a + b) R, is (A) 75 (B) 00 (C*) 5 (D) 0 Z (8 sin + 7 cos ) + i (sin + cos ) Z (sin + cos ) + i (8 sin + 7 cos ) Q. hence Z Z x iy y ix where x (8 sin + 7 cos ) and y (sin + cos ) Z Z (xy xy) + i(x + y ) 0 a 0; b x + y now, x + y (8 sin + 7 cos ) + (sin + cos ) 65 sin + 65 cos + 0 sin cos 65 + 60 sin hence Z Z 5 Ans. ] max (a) The locus of, for arg is (A) same as the locus of for arg (B) same as the locus of for arg (C*) the part of the straight line (D) the part of the straight line x y 0 with (y < 0, x > 0) x y 0 with (y > 0, x < 0) [Hint: ] (b) Let be a complex number such that arg ( ) and. Then principle value of the argument of is (A) (B*) (C) 6 (D)
arg ( ) a ray y x...() Solve i.e. x + y x and y i Hence principle value of the argument of is Ans.] Q. If & represent adjacent vertices of a regular polygon of n sides with centre at the origin and Im if Re then the value of n is equal to (A*) 8 (B) (C) 6 (D) [Hint : y x tan tan 8 5º n 60 8 5 8 y if x n ] Q.5 Let x + iy, where x, y R and i. If locus of P() satisfying Re represents a circle then maximum distance of a point on the circle from M (, ), is equal to [Note: Re() denotes the real part of.] (A) (B) 5 (C*) 6 (D) 8 We have Re x iy Re x y x + y x 0 (x ) + y. Clearly, maximum distance of M(, ) from circle is ( ) ( 0) 9 6 + 5 6. Ans.] Q.6 All real numbers x which satisfy the inequality i 5 where i, x R are (A*) [, ) (B) (, ] (C) [0, ) (D) [, 0] Q.7 Number of real or purely imaginary solution of the equation, + i 0 is : (A*) ero (B) one (C) two (D) three [Hint: Let x be the real solution. x + x i 0 x 0 & x 0 which is not possible note that the equation has no purely imaginary root as well. ] x
Q.8 A point '' moves on the curve i in an argand plane. The maximum and minimum values of are (A), (B) 6, 5 (C), (D*) 7, (x ) + i (y ) circle with centre (, ) and radius ; (, ) B C Hence OC 5 A max 5 + 7 min 5 ] O Q.9 If is a complex number satisfying the equation + i + i 8, on the complex plane then maximum value of is (A) (B*) (C) 6 (D) 8 If + i + i 8, PF + PF 8 max (B) (0, ) (0, ) P() O (0, ) (0, ) ] Q.0 Let r ( r ) be such that r r and 0 + 0 + 5 + k + + +. Then the value of k equals (A) (B) (C) (D*) We have 5 k 60 Now,,, and 5 So, k 60 60 5 0 Ans. Note for objective take ; ; ; 5 Q. Let Z be a complex number satisfying the equation (Z + ) 6 then Z has the value equal to (A) 5 / (B*) 5 / (C) 5 / (D) 5 (Z + ) 6i Z + i or i Z + i or i Z + i 5 Z 5 Z 5 / ] i Q. If i, then + 5 i + 65 is equal to (A) i (B) + i (C*) i (D) i E + 5 + 65 + 5 + ( + + ) + + ( + ) + Ans.]
Q. Consider two and as a bi a bi +, where a, b R and, where, then a bi a bi (A) Both and are purely real (B) Both and are purely imaginary (C*) is purely real and is purely imaginary (D) is purely real and is purely imaginary [Hint: Note that is real and + ( )( ) ( )( ) + ( )( ) as (given) ] r D Q. Let Z is complex satisfying the equation ( + i) + m + i 0, where m R. Suppose the equation has a real root. The additive inverse of non real root, is (A) i (B) + i (C*) i (D) Let be the real root ( + i) + m + i 0 ( + m) + i( ) 0 (real root) 6 + m 0 m. Hence, + + i (but ) + i addivitve inverse i. Ans.] Q.5 The minimum value of + i + i is (A) 5 (B) 5 (C*) (D) 5 The expression is the sum of the distance of from the two points i and + i. The minimum value is the distance between these two points 6 Ans.] [V-group 009] Q.6 The area of the triangle whose vertices are the roots of + i + i 0 is (A*) (B) 7 (C) 7 i + i + i 0 ( i) + i ( + i ) 0 ( i) [ + i ( + i)] 0 ( i) ( + i ) 0 i 8 Now, i or ( + i) Area 0 i i ±. Ans.] 0 (D) 7
Q.7 Let C and C are concentric circles of radius and 8/ respectively having centre at (, 0) on the argand plane. If the complex number satisfies the inequality, log / > then : (A*) lies outside C but inside C (B) lies inside of both C and C (C) lies outside both of C and C (D) none of these [Hint: Note that > 0, otherwise the number after log will be negative which is not possible. < ; put t (t 8) (t ) < 0 < < 8/ lies between the two concentric circles ] Q.8 The equation of the radical axis of the two circles represented by the equations, and i on the complex plane is : (A) y + 0 (B*) y 0 (C) y 0 (D) none Q.9 If - + 5i ; 5 i and is a complex number lying on the line segment joining & then arg can be : (A) (B) (C) (D*) 5 6 6 (, 5) [Hint : tan 5 > tan 5 < A/B/C cannot be the answer ] ( 5, ) / Q.0 If P and Q are represented by the and such that then the circumcentre of OPQ (where O is the origin) is (A) (B*) (C) / (D) +, We have + + 0 is purely imaginary. Hence PQR is right angled at O. Circumcentre of POQ is the mid point of PQ i.e. ( ) ]
Q. Number of such that and is (A) (B) 6 (C*) 8 (D) more than 8 Let cos x + i sin x, x [0, ). Then cos x + i sin x + cos x i sin x cos x hence cos x / or cos x / Alternatively : If cos x /, then 5 7 x, x, x, x 6 6 6 6 i If cos x, then x 5, x6, x7 5, x8 Hence there are eight solutions k cos x k + i sin x k, k,,..., 8 ] hence e i e cos cos ; e i or Q. Number of ordered pairs(s) (a, b) of real numbers such that (a + ib) 0 a ib holds good, is (A) 0 (B) 0 (C*) 0 (D) Let a + ib a ib hence we have 008 [ 0 ] 0 00 0 0 or ; if 0 0 (0, 0) if 0 0 values of Total 0. Ans.] Q. Let a point P(x, y) denoting complex number moves in argand plane satisfying 0 < Re (i) <, where i, then (A) (x, y) R 0 y (B*) (x, y) R y 0 (C) (x, y) R y (D) (x, y) R y We have 0 < Re (i ) < 0 < Re (i (x + iy)) < 0 < y < < y < 0 Option (B) is correct ] Q. Number of satisfying the relation and i i, is (A) (B*) (C) (D) We have x + y Also, i + + i A line segment between (0, ) and (0, ). So, number of solution is
i.e., i and i ] Paragraph for question nos. 5 to 7 Consider complex number and satisfying and +. Q.5 Let m and M denotes minimum and maximum value of, then (m + M) is equal to (A) 5 (B*) 6 (C) 7 (D) 8 Q.6 Re ( ) can never exceed (A) (B) (C) (D*) Q.7 If principal argument of principal argument of, then + is equal to (A) 0 (B) (C) (D*) (, 0) Im () P ( ) O (, 0) Q ( ) Re () (, 0) (, 0) (i) min m and max 5 M Hence, (m + M) + 5 6. Ans. (ii) Re ( ) Re [(cos + isin ) ] cos Re ( ) can never exceed. (iii) Clearly, 0 So, + +. Ans.] Q.8 If the complex number satisfies the condition, then the least value of (A) 5/ (B*) 8/ (C) / (D) none of these is equal to : [Hint : 8 least ]
Q.9 Let and w where, w C (where C is the set of ). If M and m respectively be the greatest and least modulus of w, then find the value of (0 m + M). Let a + ib hence a, b [, ] (a ) ib w ; a (a ) ib + b [Ans. 67] (a ) b w (a ) b w max 5 a a b b a a 5 a 5 a M, when a ( ) 5 w min m, when a ( ) 9 Hence (0 m + M) 67 + 67. Ans.] Q.0 If the expression ( + ir) is of the form of s( + i) for some real 's' where 'r' is also real and i, then the value of 'r' can be (A) cot (B*) sec (C*) 8 tan (D*) 5 tan We have ( + ri) s( + i) + ri + r i + r i s( + i) r + i(r r ) s + si r s r r Hence r r r r r r + 0 (r + ) r(r + ) 0 (r + )(r + r r) 0 r or r r + 0 r 6 r or B, C, D]