Complex Number Exercise # 5

EXERCISE # 5 Q.1 Sol. Q.2 Sol.  2i z = 2 Let z & w be two non-zero complex numbers such that |z| = | w | and Arg z + Arg w = , then z equal: (A) w (B) – w (C) w (D) – w [D] Given | z | = |  | and Arg z + Arg  =  Let  = rei |  | = r  z = | z | e i (   )  z = r. e i e i  z = (r. e i ) e i  z = (– 1) (r. e i )  z = – r. e i z=–   Option (D) is correct answer. Let z & w be two complex numbers such that |z|  1, |w|  1 and |z + i w| = |z – iw | = 2, then z equals : (A) 1 or i (B) i or – i (C) 1 or –1 (D) i or –1 [C]  iz = 1 Since, z = z z – z = 0 Imp (z) = 0 z is purely real z = x + i.0 = x | z |2  1 x2  1 – 1 < x < 1  = –   +  = 0 Re() = 0  is purely imaginary  Option (C) is correct answer. Q.3 expression ; (1  i) n1 + (1  i 3 ) n1 + (1  i 5 ) n 2 + (1  i 7 ) n 2 , where i =  1 , is a real number if : (A) n1 = n2 + 1 (B) n1 = n2 – 1 (C) n1 = n2 Given : | z | 1, |  |  1 | z + i| = |z – i  | = 2  | z + i| = |z – i  |  | z + i|2 = |z – i  |2  (z + i) (z – i  ) = (z – i  ) ( z + i)  z z + i z – iz  +   = z – i  z + iz +    i z – iz  + i  z – iz – 0  i [ z ( +  ) – z ( +  )] = 0  ( z – z) ( +  ) = 0  Either z = z or  = –  or z = z and  = –    |z + i| = 2 Squaring both sides, we get For positive integers n1, n2 the value of the Sol. (D) n1 > 0, n2 > 0 [D] (1  i) n1 + (1  i 3 ) n1 + (1  i 5 ) n 2 + (1  i 7 ) n 2 = (1  i) n1 + (1  i) n1 + (1 + i ) n 2 + (1 – i ) n 2 We have to first deal with (1  i) n1 + (1  i) n1  = 2n1 / 2 (1 / 2  i / 2 ) n1  (1 / 2  i 2 ) n1   4  4 n1  = 2n1 / 2  cos  i sin    cos = 2n1 / 2  n n (as |z|2  1, ||2  1)   = 2n1 / 2 cos 1  i sin 1  cos 2  n1 4 4 4         + i sin  2  n1  4  |z + i| = 4  z z + i z – i  z +   = 4 n 7 7  1   i sin   4 4     n 1 n 1 7 n1 7 n1   cos 4  i sin 4  cos 4  i sin 4    2  (z + i) ( z – i  ) = 4  n 1 4 Also, (1  i) n 2 + (1  i) n 2 = 2n1 / 2 . 2cos  i(z + z) = 2 40 = 2 n2 / 2 [(cos /4 + i sin /4 )  = 2 n 2 / 2 cos  = 2 Q.4 n2 7 7   +  cos  i sin  4 4   n 2 n  7 7   i sin 2  cos n 2  i sin n2  4 4 4 4   .2 cos n2 /4 Then for real value of [(1 + i ) n1 + (1 – i ) n1 + (1 + i ) n 2 + (i – i ) n 2 ], n1 > 0 & n 2 > 0  Option (D) is correct answer. n2 / 2 Find all non-zero complex numbers satisfying z = iz2. Given z = i z2 Let z = x + iy  z = x – iy We have to solve above equation 2 z = iz  x – iy = i (x + iy)2  x – iy = i (x2 – y2 + 2 ixy)  x – iy = i(x2 – y2) – 2xy Comparing real and imaginary roots  x = – 2xy and – y = x2 – y2  x + 2xy = 0 and x2 – y2 + y = 0  x (1 + 2y) and x2 – y2 + y = 0  Either x = 0 or y = – 1/2 When x = 0 0 – y2 + y = 0  y (1 – y) = 0  y = 0, 1  z = i (Non-zero) 1 1 When y = – 1/2  x2 – – =0 4 2  x2 – 3/4 = 0  x2 = 3/4  x = ± 3 /2 Sol. |z|2  + = ||2.z + z  (| z |2 + 1) = z(||2 + 1) n2 z  Im   = 0 z   Q.7 Sol. Q.8    z( z . – 1) – (  z – 1) = 0 Also, |z|2 . – ||2.z = z –  …(1) z z –   z.  = z . from (1) z(z  – 1) – (z – 1) = 0 If a1, a2, a3 .... an, A1, A2, A3 .... An, k are all real numbers, then prove that the equation imaginary roots. By theory Sol. Find the number of solution of equation |z|5 –2z |z|3 = ( z )2 |z|3 –1  z = x + iy  x, y  R & x  1. one solution. Q.9 Show that the product 2   1  i    1  i   1  1            2    2     1  i  22  1     ….   2       1  i  2n  …. 1       2     is an integer. Sol.  A12 A 22 A 2n + + .... + = k has no x  a1 x  a2 x  an 2 k n Prove that  (n  k ) cos = – where n  3 2 n k 1 For complex numbers z and w, prove that |z|2 w – | w |2 z = z – w if and only if z = w or z w = 1. | z |2 .  – |  |2 . z = z –  z z . –   .z = z –  z z – =0   Either z =  or z  = 1 Hence, Proved. n 1 Q.6  (z  – 1) (z – ) = 0  z = 3 /2 – i/2 and – 3 /2 – i/2 Hence, required non-zero complex numbers are i, 3 /2 – i/2 and – 3 /2 – i/2. Q.5 z | z |2 1 = = Purely real  |  |2 1 Sol. Q.10  1 is equal to 1  n  22 Using formula.   (1 + i) where n 2.  If cos ( – ) + cos ( – ) + cos ( – ) = – 3/2 then prove that (a) cos 2 = 0 =  sin 2 (b)  sin ( + ) = 0 =  cos ( + ) (c)  sin2 =  cos2 = 3/2 41 Sol. Q.11 Sol. Q.12 Sol. Q.13 Sol. Q.14 (d)  sin 3 = 3 sin ( +  + ) (e)  cos 3 = 3 cos ( +  +) (f) cos3 ( + ) + cos3 ( + ) + cos3( + ) = 3 cos ( + ). cos ( + ). cos ( + ) where R. Using formula. Prove that with regard to the quadratic equation z2 + (p + ip)z + q + iq = 0 where p, p, q, q are all real. (i) If the equation has one real root then q2– ppq + qp = 0 (ii) If the equation has two equal roots then p2 – p2 = 4q & pp = 2q State whether these equal roots are real or complex. Using formula. If the biquadratic x4 + ax3 + bx2 + cx + d = 0 (a, b, c, d R) has 4 non real roots, two with sum 3 + 4i & the other two with product 13 + i. Find the value of 'b'. 51 If a and b are positive integer such that N = (a + ib)3 – 107i is a positive integer. Find N. 4 Find the roots of the equation zn = (z + 1)n and show that the points which represent them are collinear on the complex plane. Hence show that these roots are also the roots of the 2 8i 16i i  4  Ki i 8i has purely imaginary value. (b) If A, B and C are the angles of a triangle D= e 2iA e iC e iB e iC e  2iB e iA e iB e iA e  2iC where i = (a) k = 3 (b) – 4 Q.17 Find all real values of the parameter a for which the equation (a –1)z4 – 4z2 + a + 2 = 0 has only pure imaginary roots. [–3, –2] Sol. Q.18 Show that the locus formed by z in the equation z3 + iz = 1 never crosses the coordinate axes in the Argand's plane. Further show that |z| = Q.19 Sol. Q.20  Im(z ) . 2 Re(z ) Im(z )  1 Dividing f(z) by z – i, we get the remainder i and dividing it by z + i, we get the remainder 1 + i. Find the remainder upon the division of f(z) by z2 + 1. iz 1  i 2 2 If Zr, r = 1, 2, 3, …. 2m, m  N are the roots of the equation Z2m + Z2m–1+ Z2m–2 +…+Z+ 1 = 0 Sol. Q.15 2m then prove that If  is the fifth root of 2 and x =  +  , prove 2 5 Z r 1 1 = – m. r 1 2 that x = 10x + 10x + 6. Sol. Q.16  1 then find the value of D. Sol. 2 m  2  m   equation  2 sin  z +  2 sin  z +1 = 0. n  n    8  i 4  3i 4i Q.21 by f(z) = |z3 – z + 2|. Find the maximum value of f(z), if |z| = 1. (a) Without expanding the determinant at any stage, find K  R such that C is the complex number, f : C  R is defined Sol. |f(z)| is maximum when z = , where  is the cube root of unity and |f(z)| = 13 42 Q.22 Sol. Q.23 Let f(x) = logcos3x (cos 2ix) if x  0 and f(0) = K Sol. 41 (where i =  1 ) is continuous at x = 0 then Q.28 If the expression z5 – 32 can be factorized into linear and quadratic factors over real coefficients as (z5 –32) = (z –2) (z2 – pz + 4) (z2 – qz + 4) then find the value of (p2 + 2p). 4 find the value of K. K = – 4/9 (i) Let Cr's denotes the combinatorial coefficients in the expansion of (1 + x)n, n  N. If the integers an = C0 + C3 + C6 + C9 + ……… bn = C1 + C4 + C7 + C10 + …….. and cn = C2 + C5 + C8 + C11 + ….. , then prove that Sol. Q.29 (a) Let z = x + iy be a complex number, where x and y are real numbers. Let A and B be the sets defined by A = {z | |z|  2} and B = {z |(1 – i)z + (1 + i) z  4}. Find the area of the region A B. (b) For all real numbers x, let the mapping 1 f(x) = , where i =  1 . If there exist x i real number a, b, c and d for which f(a), f(b), f(c) and f(d) form a square on the complex plane. Find the area of the square. Sol. (a)  – 2; (b) 1/2 (a) a 3n  b 3n  c 3n – 3an bncn = 2n, (b) (an – bn)2 + (bn – cn)2 + (cn –an)2 = 2. (ii) Prove the identity: (C0 – C2 + C4 – C6 + …..)2 + (C1 – C3 + C5 – C7 + …..)2 = 2n Sol. Q.24 Q.25 Prove that (a) cos x + nC1 cos 2x + nC2 cos 3x + …… x n 2 ..+ nCn cos (n + 1) x = 2n.cosn .cos  x 2  2  (b) sin x + nC1 sin 2x + nC2 sin 3x + ….. x n 2 ….+ nCn sin (n + 1) x = 2n.cosn .sin  x 2  2  Let A = {a  R| the equation (1 + 2i)x3 – 2(3 + i)x2 + (5 – 4i)x + 2a2 = 0}has at least one real root. Find the value of a2 . Q.30 z arg  1  z2 Q.31 aA Sol. Q.26 Q.27 18 P is a point on the Argand diagram. On the circle with OP as diameter two points Q & R are taken such that  POQ =  QOR = . If 'O' is the origin and P, Q & R are represented by the complex numbers Z1, Z2 & Z3 respectively, show that : Z22. cos 2 = Z1. Z3 cos2. 1 For x  (0, /2) and sin x = , if 3 Sol. Q.32 Sol.  sin (nx ) a  b b  3n = c then find the value of n 0 Q.33 (a + b + c), where a, b, c are positive integers. (You may use the fact that sin x = e ix  e ix ) 2i If z1, z2 are the roots of the equation az2 + bz + c = 0, with a, b, c > 0; 2b2 > 4ac > b2; z1  third quadrant; z2  second quadrant in the argand's plane then, show that Sol.   = 2 cos–1  1/ 2  b2     4ac    Find the set of points on the argand plane for which the real part of the complex number (1 + i)z2 is positive where z = x + iy, x, y  R and i =  1 . Required set is constituted by the angles without their boundaries, whose sides are the straight lines y = ( 2 –1) x and y + ( 2 +1) x = 0 containing the x-axis. Let z = 18 + 26i where z0 = x0 + iy0 (x0, y0  R) is the cube root of z having least positive argument. Find the value of x0y0 (x0 + y0). 12 Resolve z5 + 1 into linear and quadratic factors with real coefficients. Deduce that   4 sin cos = 1. 10 5 (z + 1) (z2 – 2z cos 36° + 1) (z2 – 2z cos 108° + 1) 43 Q.34 The points A,B,C depict the complex numbers z1, z2, z3 respectively on a complex plane, and numbers. Let A and B represents z1 and z2 in the angle B and C of the triangle ABC are each equal to (1/2) ( – ). Show that : (z2 – z3)2 = 4(z3 – z1) (z1 – z2) sin2 / 2 Sol. Sol. the complex plane. If AOB =  0 and OA = OB, where O is the origin prove that p2 = 4q cos2 (/2) z2 + pz + q = 0 z = Applying rotation theorem at B and C respectively, we have A(z1)  z1 =       2  2 C(z3)      2  …(1) O     2  z 2  z3 BC i  e = AC z1  z 3 (z  z ) z1  z 2 AB AC × 1 3 = × ×1 BC BC z3  z 2 z3  z 2 2  AB   (Since, |AB| = |AC|  BC  =  AB    BC  2 = … (3)   sin   sin   2  = sin(   ) / 2 = AB AC BC cos / 2 2 sin  / 2. cos  / 2  = BC AB   p  p 2  4q 2  = cos  + i sin   p  p 2  4q 2 1 (2 sin  / 2) 2 =  p  p 2  4q  p  p 2  4q – p + = cos  + i sin  p 2  4q = – p (cos + i sin ) – p 2  4q (cos + i sin ) p (cos  + i sin ) 2 sin /2 = BC/AB Put value of BC/AB = 2 sin /2 in equation (3), we get = z2 = cos  + i sin  z1  from sine rule, we get (z 3  z1 )(z1  z 2 )  2 (z1  z 2 )(z1  z 3 ) (z 2  z 3 ) 2 z2 |z | = 2 ei = cos  + i sin  z1 | z1 | …(2) Dividing equation (1) by (2), we get (z1  z 2 )(z1  z 3 ) A (z1) |OA| = |OB| | z1 | = | z2 | Apply coni method or rotation method, we have     2  i z 2  z3 z z = 2 3 e z1  z 3 z1  z 3 Q.35  p  p 2  4q B (z2) z1  z 2 AB i  e = BC z3  z 2 (z 2  z 3 ) 2 2    i  e 2  z z z1  z 2 = 1 2 z3  z 2 z3  z 2   p  p 2  4q       2  B(z2) (z 2  z 3 ) 2 1  Let roots be z2 = –  p  p 2  4q 1 4 sin 2 / 2 (z2 – z3)2 = 4 sin2 /2 (z3 – z1) (z1 – z2) Hence Proved Let z1 & z2 be roots of the equation z2 + pz + q = 0 where the coefficients p and q may be complex =–   p p  4q 2 p p 2  4q = = p 2  4q (1 + cos  + i sin ) 1  cos   i sin  1  cos   i sin  2 cos2  / 2  i 2 sin  / 2 cos  / 2 2 sin 2  / 2  i2 sin  / 2 cos  / 2 ( using cos  = 2 cos2/2 – 1 = 1 – 2 sin2 /2) 44  p p 2  4q = 2 cos  / 2(cos  / 2  i sin  / 2) i × 2 sin  / 2(sin  / 2  i cos  / 2) i  p2 × i cot  / 2(cos / 2  i sin  / 2)  = = i cot (cos / 2  i sin  / 2) 2  p p 2  4q  p2 (1 + cot2 /2) = 4q cot2 /2  p2 × cosec2 /2 = 4q cot2 /2 1 sin 2  / 2 = 4q ×  p = 4 q cos 2 Hence, Proved. 2 = i cot /2 cos2  / 2 sin 2  / 2 2 Squaring both sides, we get  p2 ( p  4q ) 2 = i2 cot2 /2 = – cot2 2  p2 = – p2 cot2 /2 + 4q cot2 /2 EXERCISE # 5 1. (D) 2. (C) 16. (a) K = 3, (b) –4 3. (D) 4. 17. [–3, –2] 19. 3 3 i i – ,– – ,i 2 2 2 2 8. one solution 12. 51 13. 4 iz 1  i 2 2 21. |f(z)| is maximum when z = , where  is the cube root of unity and |f(z)| = 13 22. K = – 4/9 25. 18 27. 41 28. 4 29. (a)  – 2; (b) 1/2 31. Required set is constituted by the angles without their boundaries, whose sides are the straight lines y = ( 2 –1) x and y + ( 2 +1) x = 0 containing the x-axis. 32. 12 33. (z + 1) (z2 – 2z cos 36° + 1) (z2 – 2z cos 108° + 1) 45