Complex Number Exercise # 1

COMPLEX NUMBER EXERCISE # 1 Question based on Q.1 Power of iota Sol. The smallest positive integer n for which Sol. 100 n n Q.5 Sol. n 100  i n ! equals (where i = = (A) –1 [C] (B) i ( x  iy ) 2 ((x  1)  iy )  ( x  1  iy ) ((x  1)  iy ) ( x 2  y 2  2i xy) [(x  1)  iy ] ( x  1) 2  y 2 always real so 2xy(x–1) – y(x2 – y2) = 0  y(2x2 – 2x – x2 + y2) = 0  y(x2 – 2x + y2) = 0 y = 0 real axis 1 ) n 0 Sol. If z2 / (z – 1) is always real, then z, can lie on(A) real axis (B) a parabola (C) imaginary axis (D) None of these [A]  = (i)n Smallest positive integer must be 2 so that (i)2 = – 1  Option (B) is correct answer. The value of 100  1 i      2  (1  i) 2  50 2 =   = i = i = –1  2  (D) 4 n 1  1  2i  =    2  Q.2 (C) 2i + 95 (D) 97 + i Q.6 If z1, z2, z3 are complex numbers such that 100  in! |z1| = |z2| = |z3| = n 0 = i0! + i1! + i2! + i3! + i4! +…..+ i100! = i + i + i2 + i6 + i4! +……+ i100! = 2i – 2 + (1 + 1 +……97 times) = 2i + 95 The value of the sum  (in + in Sol. + 1), (A) equal to 1 (B) less than 1 (C) greater than 3 (D) equal to 3 [A] Given : |z1| = |z2| = |z3| where = n 1 Sol. Question based on i =  1 , equals (A) i (B) i – 1 [B] (D) 0 Representation of complex number  1 i     2  equals- 1 1 1 =1   z1 z 2 z 3 |z1| = 1 = z1 z1 |z2| = 1 = z2 z 2 |z3| = 1 = z3 z 3  100 Q.4 (C) –i 1 1 1   z1 z 2 z 3 |z1 + z2 + z3| is- 13 Q.3 (D) – 1 50 (C) 3  (1  i) (1  i)  1 i     =    1 i   (1  i) (i  i)   (1  i) 2  =    1  1  (C) – i (B) i  1  i     2  n 1 i    = –1 is 1 i  (A) 1 (B) 2 [B] (A) 1 [D] 1 1 1   =1 z1 z 2 z 3 = 1, then   (A) 2z2 = (1 + i) z1 + (1 – i) z3 z z1 z  2  3 =1 z1z1 z 2 z 2 z 3 z 3 z1 | z1 |2  z2 | z 2 |2  z3 | z 3 |2 (B) 2z4 = (1 – i) z1 + (1 + i) z3 (C) 2z2 = (1 + i) z1 – (1 – i) z3 (D) 2z4 = (1 – i) z1 – (1 + i) z3 =1 Sol.  z1  z 2  z3 = 1 [A, B] D(z4) C(z3) A(z1) B(z2)  z1  z 2  z 3 = 1  |z1 + z2 + z3|  Option (A) is correct answer. n  1  cos   i sin     =  1  cos   i sin   Q.7  (A) cos n + i sin n (B) sin n + i cos n n n (C) cos + isin (D) cos n 2 2 [A] Sol. (1  cos   i sin ) n (1  cos   i sin ) n  2  2 cos =  2  2 cos    2i sin 2   2i sin 2   cos  2 2   cos  2 2 z1 + z3 = z2 + z4 ….(i) ABCD is a square so AD = AB e i / 2 n (z4 – z1) = (z2 – z1)i ...(ii) From (i) z4 = z1 + (z1 + z3 – z4 – z1)i  z4(1 + i) = z1 + iz3 2z4 = z1(1 – i) + z3(1 + i) Again from (i) and (ii) (z1 + z3 – z2 – z1) = z2i – z1i  z2(1 + i) = z1i + z3 2z2 = z1(1 + i) + z3(1 – i) n    n i  cos  i sin  e 2 2   2 =  n in     2 cos  i sin e   2 2  = ein = cos n + i sin n Q.8 Sol. Question based on Q.9 z1  z 3 z 2  z 4  2 2 If (x + iy)1/5 = a + ib, and u = x/a – y/b, then (A) a – b is a factor of u (B) a + b is a factor of x (C) a + ib is a factor of y (D) a – ib is a factor of a [A] (x + iy)1/5 = (a + ib)  (x + iy) + (a + ib)5 x = 5C0a5 – 5C2a3b2 + 5C4ab4 y = 5C1a4b – 5C3a2b3 + 5C5b5 x y u =  = (5C0a4 – 5C2a2b2 + 5C4b4) a b – (5C1a4 – 5C3a2b2 + 5C5b4) u = 4 (b4 – a4) = –4 (a2 + b2) (a – b) (a + b) Q.10 vertices of an equilateral triangle. If z0 be the Sol. circumcentre of the triangle then z12 + z22 + z32 = (A) z02 (B) 2z02 2 (C) 3z0 (D) 9z02 [C]  triangle is equilateral so circumcenter = z1  z 2  z 3 = z0 3  z1 + z2 + z3 = 3z0  (z1 + z2 + z3)2 = 9z02  z12  z 22  z 32 + 2(z1z2 + z2z3 + z3z1) = 9z 02  z12  z 22  z 32 = z1z2 + z2z3 + z3z1 3 (z12  z 22  z 32 )  9z 02 Algebraic operation of a complex number z1, z2, z3, z4 be the vertices A, B, C, D respectively of a square on the argand diagram taken in anticlockwise direction then Let the complex numbers z1, z2, z3 represents  z12  z 22  z 32  3z 02 Q.11 The point z1, z2 and (1 – i) z1 +iz2 of a complex plane are the vertices of a triangle which is (A) right angled isosceles Sol. (B) equilateral (C) isosceles (D) scalene [A] A(z1) Question based on Q.14 (when n is even)(A) n – 1 (B) n (C) 0 (D) None of these [C] xn – 1= 0  x = (1)1/n Given 1, 1, 2, 3, ………..n – 1 are the nth roots of unity, then. (xn – 1) = (x – 1) (x – 1) (x – 2) (x – 3) (x – 4)… .……...…….. (x – n – 1) Put x = – 1, we get [(–1)n – 1] = (–1 – 1) (–1 – 1) (–1 – 2) (–1 – 3) (–1 – 4)……..(–1 – n – 1) Since, n is even  (–2) (–1 – 1) (–1 – 2) (–1 – 3) (–1 – 4)… …....(–1 – n – 1) = 0 (1 + 1) (1 + 2) (1 + 3) (1 +4) …. (1 + n – 1) = 0  Option (C) is correct answer. C(1 – i)z1 + iz2 (z2) AB = (z2 – z1) BC = (1 – i) (z1 – z2) CA = i(z2 – z1) Clearly |AB| = |CA| angle between AB & CA is 90°. B Sol. If (3 + i) (z + z ) – (2 + i) (z – z ) + 14i = 0, then z z is equal to(A) 10 (B) 8 (C) –9 (D) –10 [A] (3 + i) (z + z ) – (2 + i) (z – z ) + 14i = 0 Let z = x + iy (3 + i) (2x) – (2 + i) (2iy) + 14i = 0  (3 + i) x – (2 + i)iy + 7i = 0  (3x + y) + (x – 2y + 7)i = 0 3x + y = 0 and x – 2y + 7 = 0 Solving we get x = –1, y = 3 zz = x2 + y2 = 1 + 9 = 10 Q.15 Sol.  If arg (z) < 0, then arg (– z) – arg (z) =   (A)  (B) –  (C) – (D) 2 2 [A] Given Arg (z) < 0 i.e. Arg (z) = –  Then = r [(cos (– ) + i sin (– )] z = r [cos  – i sin ] – z = – r [cos  – i sin ] – z = – r cos  + i r sin  – z = r[– cos  + i sin ] – z = r [cos ( – ) + i sin ( – )]  Arg ( – z) =  –   Arg (– z) – Arg (z) =  –  – (– ) =–+ =  Option (A) is correct answer. If 1, 2, 3,... 8 are the 8, 8th roots of unity then (A) (1)3 + (2)3 + (3)3 +...+ (8)3 = 8 (B) 1 + 2 + 3 +... + 8 = 1 (C) (1)50 + (2)50 ...+ (8)50 = 0 (D) (1)16 + (2)16 ...+ (8)16 = 0 Sol. Q.13 If 1, 1, 2,..... n–1 are the nth roots of unity, then (1 + 1) (1 + 2) .......(1 + n–1) is equal to Sol. Q.12 De Moiver's Theorem & roots of unity [C] x8 – 1 = (x – 1) (x – 2) (x – 3) (x – 4)….. …….. (x – 8) 18  1; 82  1; 83  1; 84  1;.........; 88  1 We have to check for every option as follows. (1)3 + (2)3 +(3)3 +(4)3 + …+ (8)3 = (1 + 2 + 3 + …+ 8)3 – 61 2 3 – 31 2 =0–0–0 =0  Option (A) is not correct answer. For B : Sum of roots, 1 + 2 + 3 + ….8 = 0   Option (B) is not correct answer. For C : (1)50 + (2)50 +(3)50 +(4)50 +…. + (8)50 = (18)6 12 + (28)6  22 + (38)6  32 + (48)6  24 + ….. + (88)6  82 = 12   22  32   24  .......82 = (1 + 2 + 3 + 4 + ……….. + 8)2 – 212 =0–0 =0  Option (C) is correct answer. For D : (1)16 + (2)16 + (3)16 + (4)16 + ……. + (8)16 = (18)2 + (28)2 + (38)2 + (48)2 + …… + (88)2 =1+1+1+1+1+1+1+1 =8 Since, 18 = 1;  82 = 1 ………. ;  88 = 1  (i101 + z101)103 = [ i101 + i101202]103 = (i + i)103 = [– i2]103 = –i3 206 = i2 = z  If 1, , 2,......n – 1 are the nth roots of unity, then (2 – ) (2 – 2).....(2 – n–1) equals (A) 2n – 1 (B) nC1 + nC2 +...+nCn Sol. (–64)1/4 equals(A) ± 2(1 + i) (B) ± 2 (1 – i) (C) ± 2(1 ± i) (D) None of these [C] (– 64)1/4 = [(– 64)1/2]1/2 = (± 8i)1/2 = [4 (± 2i)]1/2 = [4 (1 ± 2i – 1)]1/2 = [4 (1 ± i)2]1/2 = ± 2 (1 ± i) (C) [2n+1C0 + 2n+1C1 +....+ 2n+1Cn]1/2 –1 Q.19 If  is non real and  = Q.18  Option (D) is not correct answer. Q.16 Sol.  1 i 3  3 i  = i 2 = – i  2  2   z= (D) none of these [A,B,C] Let equation xn – 1 = 0 have n roots as (xn – 1) = (x – 1) (x – ) (x – 2) (x – 3) …. …. (x – n – 1)  |1    2   2  1| 2|1 x 1 = (x – ) (x – 2) (x – 3) …. (x – n – x 1 n = 2|2 ) Put x = 2, we get 2 = 2 =4 2 1 = 2n – 1 2 1 n 3 2 2 n 1 (2n + 1)  = C1 + (2n + 1)C2 + … + 2n + 1Cn 2 (2n + 1)C1 + (2n + 1)C2 + … + (2n + 1)Cn = 22n  ((2n + 1)C1 + (2n + 1)C1 + ………. + (2n + 1)Cn)1/2 – 1 = (22n)1/2 – 1  Option (A), (B) and (C) are correct answer. Q.17 Sol. 3 i , then (i101 + z101)103 equals2 (A) iz (B) z (C) z (D) None of these [B] If z = 4 2 |   3  4 | 1  ( || = 1 ) 6 Q.20 The value of Sol. (A) –1 (C) –i [D] 6  cos k 1    sin k 1 –i ( 5 = 1) [ 1 +  + 2 + 3 + 4 = 0] n–1 = (2 – ) (2 –  ) (2 –  ) ………. (2 –  ) Also, (1 + 1)n = 2n = nC0 + nC1 + nC2 + nC3 + … + nCn  nC1 + nC2 + nC3 + …. + nCn = 2n – 1 Also, (1 + 1)2n + 1 = (2n + 1)C0 + (2n + 1)C1 + (2n + 1) C2 + (2n + 1)C3 + ………. + (2n + 1)Cn + 2n + 1Cn + 1 + ...+ (2n + 1)C2n + 1  22n + 1 = 2 × (2n + 1C1 + 2n + 1C2 + ….. + (2n + 1)Cn) 2   2  1| = 2|1    1  2 1 then the value of is equal to(B) 2 (D) none of these 2 (A) 4 (C) 1 [A] Sol. 5 2k 2k   i cos  is7 7  (B) 0 (D) i 2k 2k  i sin 7 7  6 2k 2k   2k 2k    i sin  i sin –i  cos      cos 7 7   7 7  k 0   k 0 sum of all seven values of seventh root of unity = 0 –i{0 – (cos 0 + i sin 0)} = – i{–1 + 0} = i  Q.21 The value of the expression 1. (2 – ) . (2 – 2) + 2 . (3 – ) (3 – 2) + ....... + (n – 1) (n – ) (n – 2), where  is an imaginary cube root of unity is n ( n  1)  (A)   2   2 2  n (n  1)  (B)   –n  2  (C) ellipse 2 Sol.  n ( n  1)  (C)   + n (D) None of above 2   [B] General term (r – 1) (r – ) (r – 2) (r – 1) {r2 – r2 – r+ 3} (r – 1) {r2 – r( + 2) + 3} (r – 1) (1 + r + r2) = 3 – 1 n  (r 3  1)  r 1 Q.22 Sol. n  r3  r 1 Sol.  z=x+i  1  1 x + iy = cos   r   + isin  r   r r    2  1 x = cos   r   r  If  (1) is a cube root of unity and (1 + )7 = A + B, then A & B are respectively the numbers(A) 0, 1 (B) 1, 1 (C) 1, 0 (D) –1, 1 [B]  is a cube roots of unity i.e.  =  1 i 3 but  1 2  1 y = sin   r   r  x2 1  r   r  Q.24  1 i 3 =– 2 = 1 i 3 2 = 1 i 3 1 1 + – 2 2 2 =1+ Question based on Q.23 1 and z1 is any point on a fixed z1 circle with the centre at the origin then z lies on (A) straight line (B) circle  1 r   r  2 1 If z  C & log|z+i| |3 + 4i| < log|z+i| |5 +12i|, (B) z lies outside a circle passing through the origin (C) z lies inside the circle |z + i| < 5 (D) z lies outside the curve z = – i + e i,   R. Sol. [D] Given, z  C log |z + i| |3 + 4i | < log | z + i | | 5 + 12i |  above inequality holds only if |z + i| > 1  |z + i| > |ei|  z + i > ei  z > – i + ei i.e. z lies outside the curve z = – i + ei  Option (D) is correct answer. Q.25 If z0, z1 represent points P, Q on the locus |z – 1| = 1 and the line segment PQ subtends an angle /2 at the point z = 1 then z1 is equal to- Geometrical figures in complex plane If z = z1 + y2 (A) inside a circle passing through the origin  1 i 3 2 =1+ = A + .B  A = 1 and B = 1  Option (B) is correct answer. 2  then z lies-  1 i 3  1 i 3 Let  =  2 = 2 2 1 +  + 2 = 0  1 +  = – 2  (1 + )7 = (– 2)7 = – 14 = – 12 × 2 = –2 [C] Let z1 = r cos  + irsin 1 1 z = z1 + = r(cos + isin) + (cos – r z1 isin)  n (n  1)  1   n  2  r 1 n (D) hyperbola Sol. i z0 1 (A) 1 + i (z0 – 1) (B) (C) 1 – i (z0 + 1) (D) i (z0 – 1) [A, C] Q z1 /2 Q z0 Pz 0 (1, 0) /2 1 Q.26 (D) None of these Sol. [C] |z–1|+|z+1|4 Let P = x + iy, A = 1 + i0 and B = –1 + i0 then PA + PB 4 (constant) Clearly locus of z is interior and boundary of an ellipse. Q.28 Among the complex numbers z, satisfying the condition |z + 1 – i|  1, the number having the least positive argument is- (1, 0) (z1 – 1) = (z0 – 1) e i / 2 z1 – 1 = (z0 – 1) i z1 = 1 + i (z0 – 1) (z0 – 1) = (z1 – 1) e (z0 – 1) = (z1 – 1) i z1 – 1 = –i (z0 – 1) z1 = 1 – i (z0 – 1) (C) interior and to boundary of an ellipse Pz i / 2 Locus of the point z satisfying the equation |iz – 1| + |z – i| = 2 is- Sol. (A) straight line segment (B) a circle (A) 1 – i (B) –1 + i (C) –i (D) None of these [D] |z + 1 – i|  1 |x + iy + 1 – i|  1 (x + 1)2 + (y – 1)2  1 (–1,1) (C) an ellipse A(0,1) (D) a pair of straight lines Sol. [A] | iz – 1 | + | z – i | = 2 | iz + i2 | + | z – i | = 2 |i||z+i|+|z–i|=2 |z+i|+|z–i|=2 Let z = x + iy  | x + iy + i | + | x + iy – i | = 2  | x + i(y + 1) | + | x + i(y – 1) | = 2  x 2  ( y  1) 2 + point A(0, 1) = i having least positive argument Q.29 x 2  ( y  1) 2 = 2  x 2  ( y  1) 2 = 2 – x 2  ( y  1) 2 Squaring both sides, we have  x2 + (y + 1)2 = 4 + x2 + (y – 1)2 – 4 x 2  ( y  1) 2  y2 + 1 + 2y = 4 + y2 + 1 – 2y – 4 x 2  ( y  1) 2  4y = 4 – 4 x 2  ( y  1) 2  (y – 1) = – x 2  ( y  1) 2   Q.27  (y – 1)2 = x2 + (y – 1)2  x2 = 0 x=0 Which is a straight line  Option (A) is correct answer. The region of argand plane defined by |z – 1| + |z + 1|  4 is(A) interior of an ellipse (B) exterior of a circle The vector z = –4 + 5i turned counter stretched 1.5 times. The complex number corresponding to the newly obtained vector is15 15 (A) 6 – i (B) – 6 + i 2 2 (C) 6 + Sol. 15 i 2 (D) none of these [A] 3 z ei 2 3 = (– 4 + 5i) (cos  + i sin ) 2 3 = – (– 4 + 5i) 2 15 =6– i 2 z1 =    Fill in the blanks type questions Q.30 The set of points in an Argand diagram which satisfy both |z|  4 and arg (z) = /3 is Sol. |z|  4 locus of circle A (z1) (0,1) locus of z /3  equation of straight line 3 radius of circle Q.31 Sol. (AB)2 = (AC)2 ….(1) (AC)2 = (AB)2 2 2 2 a +1=b +1 a + 1 = (a – 1)2 + (b – 1)2 a=b a=b a2 – 4a + 1 = 0 If m and x are two real numbers, then e2mi cot e 2micot 1  xi  1     xi  1  x  1  xi      1  xi  = e 2micot = e 2micot = e 2micot = e × m x 1 1 is equal to – m (i) ( x  i) m x C (z3) (0,0) (z2) B (1,b) arg (z) = m same sign a=2± 3 a = 2 3 a=2+ 3 b = 2 3 a=2– 3 (0 < a < 1) (i) m ( x  i) m 1 1 x x i   x    2 1 x2  1 x m i   x    2 1 x2  1 x m Q.34 (cos(cot1 x )  i sin(cot 1 x )) m (cos cot 1 x  i sin cot 1 x ) m 2mi cot 1 x –1 ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the points D and M represent the complex numbers 1+i and 1–i respectively, then A represents the complex number........... or........ Sol. –1 (cos (m cot x) – i sin (m cot x) C (z3) (z4) D (cos (m cot–1 x) – i sin (m cos–1 x) = e 2micot = e 2mi cot 1 x 1 x z [cos2(m cot–1x) – sin2 (m cot–1x) – i2 sin (m cot–1 x) cos (m cot–1 x)] –1 (z1) A [cos (2m cot x) – i sin (2m cot x)] = e 2micot =1 1 x × e 2micot 1 Let point A = z1 (z 4  z) (z1  z) i / 2  e (z 4  z) (z1  z) x Q.32 For any two complex numbers z1, z2 and any real number a and b, |az1–bz2|2 + |bz1 + az2|2 = ........ Sol. |az1 – bz2|2 + |bz1 + az2|2 (az1 – bz2) (az1  bz 2 ) + (bz1 + az2) (bz1  az 2 ) 2(z4 – z) = (z1 – z) (±i) z4 = (1 + i) put z = 3 – i/2 z=1–i z = 1 – (3/2) i Q.35 a2|z1|2 – abz1 z 2 – ab z1 z2 + b2 |z2|2 + b2 |z1|2 + 2 (a2 + b2) (|z1|2 + |z2|2) If  are the numbers between 0 & 1 such that points z1 = a + i, z2 = 1 + bi & z3 = 0 form an equilateral triangle, then a = ........ & b = ......... Sol. Suppose z1, z2, z3 are the vertices of an equilateral triangle inscribed in the circle 2 abz1 z 2 + ab z1 z2 + a |z2| Q.33 B (z2) –1 |z| = 2. If z1 = 1 + i 3 then z2 = ........, z3 =… Sol. z2 Principal arg (z) = – z1 120° O 101° 120° z2 = z1 e i 2 3  1 i 3  = (1 + i 3 )     2  2   If log1/2 Sol. log1/2  (i 3 ) 2  1  3  1   2 = 2 2 z3 = z1 Q.36 Sol. 4 i e 3 z + iz y | z  1 | 4 >1 3 | z  1 | 2 | z 1 |  4 1  3 | z  1 | 2 2  2| z – 1 | + 8 < 3| z – 1 | – 2  | z – 1 | > 10  z lies exterior of the circle with center 1 + i0 and radius 10 = 1 i 3 The area of the triangle on the Argand diagram formed by the complex numbers z, iz and z + iz is ………….. We have ; iz = zei/2. This implies that iz is the vector obtained by rotating vector z in anticlockwise direction through 90º. Therefore, OA AB. So, | z  1 | 4 >1 then locus of z is 3 | z  1 | 2 exterior to circle with center 1 + i0 & radius 10. Q.38 z3 Q.39 Sol. The cube roots of unity when represented on Argand diagram form the vertices of an equilateral triangle. 1 = ei0  = ei 2 / 3 2 = e i 4  / 3  ei 2 / 3  B equilateral  iz x' O  2 A x 1 (0,0) 1ei0 y' 1 OA×OB 2 1 1 = |z| |iz| = |z|2. 2 2 ei4/3 2 Area of OAB =  True or false type questions Q.37 Sol. If z = 5  12i  5  12i 5  12i  5  12i  of argument z is . 2 5  12i  5  12i z= 5  12i  5  12i = , then principal value ( 5  12i  5  12i ) 2 (5  12i)  (5  12i) 5  12i  5  12i  2 25  144 24i 36 3 =  i 24i 2 = Q.40 Sol. centre of circle is (0, 0) Radius is ‘1’ Let P(x) and Q(x) be two polynomials. Suppose that f(x) = P(x3) + xQ(x3) is divisible by x2 + x + 1, then f(x) is divisible by (x – 1). Given : f(x) = P(x3) + x.Q(x3) Since, f(x) is divisible by (x2 + x + 1) i.e. It must be divisible by (x – 1). Because (x2 + x + 1) is a factor of (x3 – 1) = (x – 1) (x2 + x + 1) Hence, Both P(x) and Q(x) must be divisible by (x – 1) ANSWER KEY EXERCISE # 1 Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B C B D A A A A A,B C A A A C C Q.No. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 C B A D B B C D A,C A C D A Ans. A,B,C 30. a radius of a circle 34. 3 – i 3 or 1 – i 2 2 40. True 31. 1 32. (a2 + b2) (|z1|2 +|z2|2) 35. –2, 1 –i 3 36. | z |2 2 37. False 33. 2 – 38. True 3,2– 3 39. True