Complex Numbers JEE Mathematics Worksheets Set 01

Subjective Questions

Question. Find the real values of x and y for which the following equation is satisfied
\( \frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i \)
Answer: \( \frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i \)
\( (3 - i) [x + i(x - 2)] + [2y + i(1 - 3y)] (3 + i) = 10i \)
\( (3x + x - 2) + i[3x - 6 - x] + (6y - 1 + 3y) + i(2y + 3 - 9y) = 10i \)
\( (4x + 9y - 3) + i[2x - 7y - 3] = 10i \)
By comparing
\( x = 3, y = -1 \)

Question. Find the square root of
(i) \( 7 + 24i \)
(ii) \( 4 + 3i \)
Answer: (i) Let \( \sqrt{7 + 24i} = x + iy \)
\( 7 + 24i = x^2 - y^2 + 2ixy \)
\( x^2 - y^2 = 7 \) ....(1)
\( xy = 12 \) ....(2)
\( (x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2 = 49 + 576 = 625 \)
\( x^2 + y^2 = \pm 25 \)
\( x^2 + y^2 = 25 \) ...(3)
\( x^2 - y^2 = -25 \) (reject)
From (1) and (3)
\( 2x^2 = 32 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4 \)
\( x = -4 \Rightarrow y = -3 \)
\( \sqrt{7 + 24i} = \pm (4 + 3i) \)
(ii) Let \( \sqrt{4 + 3i} = x + iy \)
\( 4 + 3i = x^2 - y^2 + 2ixy \)
\( x^2 - y^2 = 4 \)
\( 2xy = 3 \)
\( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 16 + 9 = 25 \)
\( x^2 + y^2 = 5 \)
\( x^2 = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}} \)
\( y^2 = \frac{1}{2} \Rightarrow y = \pm \frac{1}{\sqrt{2}} \)
\( x = \frac{3}{\sqrt{2}} \Rightarrow y = \frac{1}{\sqrt{2}} \)
\( x = -\frac{3}{\sqrt{2}} \Rightarrow y = -\frac{1}{\sqrt{2}} \)
\( \sqrt{4 + 3i} = \pm \frac{1}{\sqrt{2}} (3 + i) \)

Question. If \( |z_1| = |z_2| = \dots = |z_n| = 1 \) then show that
(i) \( \overline{z}_1 = \frac{1}{z_1} \)
(ii) \( |z_1 + z_2 + \dots + z_n| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right| \)
And hence interpret that the centroid of polygon with 2n vertices \( z_1, z_2, \dots, z_n, \frac{1}{z_1}, \frac{1}{z_2}, \dots, \frac{1}{z_n} \) (need not be in order) lies on real axis.
Answer: \( |z_1| = |z_2| = |z_3| = \dots = |z_n| = 1 \)
(i) \( |z_1| = 1 \)
\( |z_1|^2 = 1 \)
\( z_1 \overline{z}_1 = 1 \Rightarrow \overline{z}_1 = \frac{1}{z_1} \)
(ii) \( \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right| \)
\( = |\overline{z}_1 + \overline{z}_2 + \dots + \overline{z}_n| \)
\( = |\overline{z_1 + z_2 + \dots + z_n}| \)
\( = |z_1 + z_2 + \dots + z_n| = \text{LHS} \quad \text{H.P.} \)
\( \because |z_1| = 1 \Rightarrow \left| \frac{1}{z_1} \right| = 1 \)
\( |z_2| = 1 \Rightarrow \left| \frac{1}{z_2} \right| = 1 \)
\( |z_n| = 1 \Rightarrow \left| \frac{1}{z_n} \right| = 1 \)
Hence that 2n points are the vertices of a regular polygon.

Question. If n is a positive integer, prove the following
(i) \( (1 + \cos \theta + i \sin \theta)^n + (1 + \cos \theta - i \sin \theta)^n = 2^{n+1} \cos^n \frac{\theta}{2} \cos \frac{n\theta}{2} \).
(ii) \( (1 + i)^n + (1 - i)^n = 2^{\frac{n}{2} + 1} \cos \frac{n\pi}{4} \)
Answer: (i) \( (1 + \cos \theta + i \sin \theta)^n + (1 + \cos \theta - i \sin \theta)^n \)
\( = \left( 2 \cos^2 \frac{\theta}{2} + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right)^n + \left( 2 \cos^2 \frac{\theta}{2} - i 2 \cos \frac{\theta}{2} \sin \frac{\theta}{2} \right)^n \)
\( = 2^n \cos^n \frac{\theta}{2} \left[ \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right)^n + \left( \cos \frac{\theta}{2} - i \sin \frac{\theta}{2} \right)^n \right] \)
\( = 2^n \cos^n \frac{\theta}{2} \left[ \cos \frac{n\theta}{2} + i \sin \frac{n\theta}{2} + \cos \frac{n\theta}{2} - i \sin \frac{n\theta}{2} \right] \)
\( = 2^{n+1} \cos^n \frac{\theta}{2} \cos \frac{n\theta}{2} = \text{RHS} \)
(ii) \( (1 + i)^n + (1 - i)^n \)
\( = (\sqrt{2} e^{i \frac{\pi}{4}})^n + (\sqrt{2} e^{-i \frac{\pi}{4}})^n \)
\( = 2^{\frac{n}{2}} [ e^{i \frac{n\pi}{4}} + e^{-i \frac{n\pi}{4}} ] \)
\( = 2^{\frac{n}{2} + 1} \cos \frac{n\pi}{4} \)

Question. Find the values(s) of the following
(i) \( \left( \frac{1}{2} + \frac{\sqrt{-3}}{2} \right)^3 \)
(ii) \( \left( \frac{1}{2} + \frac{\sqrt{-3}}{2} \right)^{3/4} \)
Hence find continued product if two or more distinct values exists.
Answer: (i) \( \left( \frac{1}{2} + \frac{\sqrt{-3}}{2} \right)^3 = \left( \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^3 = -(-\omega^2)^3 = -(\omega^3)^2 = -1 \)
(ii) \( \left( \frac{1}{2} + \frac{\sqrt{-3}}{2} \right)^{3/4} = \left( \frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{3/4} \)
\( = \left[ e^{i \left( 2n\pi + \frac{\pi}{3} \right)} \right]^{3/4} \)
\( = e^{i(6n + 1)\frac{\pi}{4}} \) where \( n = 0, 1, 2, 3 \)
Product = \( -1 \)

Question. Let \( \text{I} : \text{Arg} \left( \frac{z - 8i}{z + 6} \right) = \pm \frac{\pi}{2} \)
\( \text{II} : \text{Re} \left( \frac{z - 8i}{z + 6} \right) = 0 \)
Show that locus of z in I or II lies on \( x^2 + y^2 + 6x - 8y = 0 \)
Hence show that locus of z can also be represented by \( \frac{z - 8i}{z + 6} + \frac{\overline{z} + 8i}{\overline{z} + 6} = 0 \). Further if locus of z is expressed as \( |z + 3 - 4i| = R \), then find R.
Answer: \( \text{I} : \text{Arg} \left( \frac{z - 8i}{z + 6} \right) = \pm \frac{\pi}{2} \)
I represent a circle with diameter ends are (-6, 0) & (0, 8)
Hence equation of circle
\( (x + 6) x + y (y - 8) = 0 \)
\( x^2 + 6x + y^2 - 8y = 0 \)
\( \text{II} : \text{Re} \left( \frac{z - 8i}{z + 6} \right) = 0 \)
\( \text{Re} \left( \frac{x + i(y - 8)}{(x + 6) + iy} \times \frac{(x + 6) - iy}{(x + 6) - iy} \right) = 0 \)
Hence I and II represent same circle
\( \frac{z - 8i}{z + 6} + \frac{\overline{z} + 8i}{\overline{z} + 6} = 0 \)
\( (z - 8i) (\overline{z} + 6) + (z + 6) (\overline{z} + 8i) = 0 \)
\( z\overline{z} + z(3 + 4i) + \overline{z}(3 - 4i) = 0 \)
It is also represent +ve same circle
\( |z + 3 - 4i| = R \)
\( (x + 3)^2 + (y - 4)^2 = R^2 \)
\( x^2 + 6x + y^2 - 8y + 9 + 16 = R^2 \)
\( \Rightarrow R^2 = 25 \Rightarrow R = 5 \)

Question. If \( \alpha \) is imaginary \( n^{\text{th}} \) \( (n \ge 3) \) root of unity then show that \( \sum_{r=1}^{n-1} (n - r)\alpha^r = \frac{n\alpha}{1 - \alpha} \). Hence deduce that \( \sum_{r=1}^{n-1} (n - r)\sin \frac{2r\pi}{n} = \frac{n}{2} \cot \frac{\pi}{n} \)
Answer: \( \sum_{r=1}^{n-1} (n - r)\alpha^r \)
\( = (n - 1) \alpha + (n - 2) \alpha^2 + (n - 3) \alpha^3 + \dots + \{n - (n - 1)\} \alpha^{n-1} \)
\( = n [\alpha + \alpha^2 + \alpha^3 + \dots + \alpha^{n-1}] - [\alpha + 2\alpha^2 + 3\alpha^3 + \dots + (n - 1)\alpha^{n-1}] \)
\( S = -n - S_1 \) ....(1)
\( S_1 = \alpha + 2\alpha^2 + 3\alpha^3 + \dots + (n - 1) \alpha^{n-1} \)
\( \alpha S_1 = \alpha^2 + 2\alpha^3 + \dots + (n - 2) \alpha^{n-1} + (n - 1) \alpha^n \)
\( S_1 (1 - \alpha) = \alpha + \alpha^2 + \alpha^3 + \dots + \alpha^{n-1} - (n - 1)\alpha^n \)
\( S_1 (1 - \alpha) = -1 - n\alpha^n + \alpha^n \)
\( S_1 = \frac{-n\alpha^n}{1 - \alpha} \)
\( S = -n + \frac{n\alpha^n}{1 - \alpha} = -n + \frac{n}{1 - \alpha} = \frac{-n + n\alpha + n}{1 - \alpha} = \frac{n\alpha}{1 - \alpha} \)

Question. If \( \alpha, \beta, \gamma \) are roots of \( x^3 - 3x^2 + 3x + 7 = 0 \) and \( \omega \) is imaginary cube root of unity, then find the value of \( \frac{\alpha - 1}{\beta - 1} + \frac{\beta - 1}{\gamma - 1} + \frac{\gamma - 1}{\alpha - 1} \).
Answer: \( x^3 - 3x^2 + 3x + 7 = 0 \)
\( (x - 1)^3 = -8 \)
\( \frac{x - 1}{-2} = (1)^{1/3} = 1, \omega, \omega^2 \)
\( \alpha = -1, \beta = 1 - 2\omega, \gamma = 1 - 2\omega^2 \)
\( \frac{\alpha - 1}{\beta - 1} + \frac{\beta - 1}{\gamma - 1} + \frac{\gamma - 1}{\alpha - 1} \)
\( = \frac{-2}{-2\omega} + \left( \frac{-2\omega}{-2\omega^2} \right) + \left( \frac{-2\omega^2}{-2} \right) \)
\( = \frac{1}{\omega} + \frac{1}{\omega} + \omega^2 = \omega^2 + \omega^2 + \omega^2 = 3\omega^2 \)

Question. Given \( z_1 + z_2 + z_3 = A \), \( z_1 + z_2 \omega + z_3 \omega^2 = B \), \( z_1 + z_2 \omega^2 + z_3 \omega = C \), where \( \omega \) is cube root of unity,
(a) express \( z_1, z_2, z_3 \) in terms of A, B, C.
(b) prove that, \( |A|^2+|B|^2+|C|^2=(|z_1|^2+|z_2|^2+|z_3|^2) \).
(c) prove that \( A^3 + B^3 + C^3 - 3ABC = 27z_1z_2z_3 \)
Answer: \( z_1 + z_2 + z_3 = A \), \( z_1 + z_2\omega + z_3\omega^2 = B \), \( z_1 + z_2\omega^2 + z_3\omega = C \)
(a) adding (1), (2) and (3)
\( 3z_1 + z_2 (1 + \omega + \omega^2) + z_3 (1 + \omega + \omega^2) = A + B + C \)
\( \Rightarrow z_1 = \frac{A + B + C}{3} \)
\( (1) + \omega^2(2) + \omega(3) \)
\( z_1 (1 + \omega + \omega^2) + 3z_2 + z_3 (1 + \omega + \omega^2) = A + B\omega^2 + C\omega \Rightarrow z_2 = \frac{A + B\omega^2 + C\omega}{3} \)
Similarly \( z_3 = \frac{A + B\omega + C\omega^2}{3} \)
(b) \( |A|^2 + |B|^2 + |C|^2 = 3(|z_1|^2 + |z_2|^2 + |z_3|^2) \)
\( = |z_1|^2 + |z_2|^2 + |z_3|^2 + z_2 \overline{z}_1 + z_3 \overline{z}_1 + z_1 \overline{z}_2 + z_3 \overline{z}_2 + z_1 \overline{z}_3 + z_2 \overline{z}_3 \)
Similarly for \( |B|^2 \) & \( |C|^2 \) and add then
\( |A|^2 + |B|^2 + |C|^2 = 3 (|z_1|^2 + |z_2|^2 + |z_3|^2) \)
as \( (1 + \omega + \omega^2 = 0) \)
(c) multiple \( z_1, z_2 \) & \( z_3 \) from part (a)
\( A^3 + B^3 + C^3 - 3ABC = 27 z_1z_2z_3 \)

Question. Prove that, with regard to the quadratic equation \( z^2 + (p + ip') z + q + iq' = 0 \); where \( p, p', q, q' \) are all real.
(a) If the equation has one real root then
\( q'^2 - pp' q' + pq'^2 = 0 \).
(b) If the equation has two equal roots then
\( p^2 - p'^2 = 4q \) & \( pp' = 2q' \).
State whether these equal roots are real or complex.
Answer: \( z^2 + (P + ip') z + q + iq' = 0 \)
(a) If the equation has one real root
put \( z = x \)
\( x^2 + (P + ip') x + q + iq' = 0 \)
\( x^2 + px + q + i (p'x + q') = 0 \)
\( x^2 + px + q = 0 \quad p'x + q' = 0 \)
\( x^2 + px + q = 0 \quad x = -\frac{q'}{p'} \)
\( \left(-\frac{q'}{p'}\right)^2 - p\left(\frac{q'}{p'}\right) + q = 0 \)
\( q'^2 - pp'q' + qp'^2 = 0 \) H.P.
(b) Let roots be \( \alpha \) & \( \beta \)
\( \alpha + \beta = -(P + iP') \)
\( \alpha\beta = q + iq' \quad \therefore \alpha = \beta \)
\( 2\alpha = -(p + ip') \)
\( \alpha^2 = (q + iq') \)
\( 4\alpha^2 = p^2 - p'^2 + 2ipp' \)
\( 4(q + iq') = p^2 - p'^2 + 2ipp' \)
\( 4q = p^2 - p'^2 \quad \& \quad 2q' = pp' \)

Question. Simplify and express the result in the form of a+bi
(a) \( \left(\frac{1 + 2i}{2 + i}\right)^2 \)
(b) \( -i (9 + 6i) (2 - i)^{-1} \)
(c) \( \left(\frac{4i^3 - i}{2i + 1}\right)^2 \)
(d) \( \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \)
(e) \( \frac{(2 + i)^2}{2 - i} - \frac{(2 - i)^2}{2 + i} \)
Answer: (a) \( \left(\frac{1 + 2i}{2 + i}\right)^2 = \left(\frac{(1 + 2i)(2 - i)}{5}\right)^2 \)
\( = \frac{1}{25} [2 - i + 4i + 2]^2 = \frac{1}{25} [4 + 3i]^2 \)
\( = \frac{1}{25} [16 - 9 + 24i] = \frac{1}{25} [7 + 24i] \)
(b) \( -i (9 + 6i) (2 - i)^{-1} = -\left( \frac{9i - 6}{2 - i} \right) = \frac{6 - 9i}{2 - i} \)
\( = \frac{(6 - 9i)(2 + i)}{5} = \frac{12 + 6i - 18i + 9}{5} = \frac{21 - 12i}{5} \)
(c) \( \left(\frac{4i^3 - i}{2i + 1}\right)^2 = \left(\frac{-4i - i}{2i + 1}\right)^2 = \left(\frac{-5i(2i - 1)}{5}\right)^2 \)
\( = (-i (2i - 1))^2 = 3 + 4i \)
(d) \( \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \)
\( = \frac{(3 + 2i)(2 + 5i)}{29} + \frac{(3 - 2i)(2 - 5i)}{29} \)
\( = \frac{1}{29} [6 + 15i + 4i - 10 + 6 - 15i - 4i - 10] \)
\( = \frac{-8}{29} + 0i \)
(e) \( \frac{(2 + i)^2}{2 - i} - \frac{(2 - i)^2}{2 + i} \)
\( = \frac{1}{5} [(2 + i)^3 - (2 - i)^3] \)
\( = \frac{1}{5} [8 - i + 6i(2 + i) - (8 + i - 6i(2 - i))] \)
\( = \frac{1}{5} [8 - i + 12i - 6 - 8 - i + 12i + 6] \)
\( = \frac{22}{5} i \)

Question. Given that \( x, y \in \mathbb{R} \) solve
(a) \( (x + 2y) + i(2x - 3y) = 5 - 4i \)
(b) \( (x + iy) + (7 - 5i) = 9 + 4i \)
(c) \( x^2 - y^2 - i(2x + y) = 2i \)
(d) \( (2 + 3i)x^2 - (3 - 2i) y = 2x - 3y + 5i \)
(e) \( 4x^2 + 3xy + (2xy - 3x^2)i = 4y^2 - (x^2/2) + (3xy - 2y^2) i \)
Answer: (a) \( (x + 2y) + i(2x - 3y) = 5 - 4i \)
\( x + 2y = 5 \)
\( 2x - 3y = -4 \)
By solving \( x = 1 \); \( y = 2 \)
(b) \( (x + iy) + (7 - 5i) = 9 + 4i \)
\( x + 7 + i(y - 5) = 9 + 4i \)
\( x + 7 = 9 \Rightarrow x = 2 \)
\( y - 5 = 4 \Rightarrow y = 9 \)
(c) \( x^2 - y^2 - i(2x + y) = 2i \)
\( x^2 - y^2 = 0 \Rightarrow y = \pm x \)
\( y = x \Rightarrow x = -\frac{2}{3} \Rightarrow y = -\frac{2}{3} \)
\( y = -x \Rightarrow x = -2 \Rightarrow y = -2 \)
points \( \left(-\frac{2}{3}, -\frac{2}{3}\right), (-2, -2) \)
(d) \( (2 + 3i) x^2 - (3 - 2i) y = 2x - 3y + 5i \)
\( (2x^2 - 3y) + i (3x^2 + 2y) = 2x - 3y + 5i \)
\( 2x^2 - 3y = 2x - 3y \) & \( 3x^2 + 2y = 5 \)
\( 2x^2 = 2x \Rightarrow x = 0, 1 \)
\( x = 0 \Rightarrow y = \frac{5}{2} \) points \( \left(0, \frac{5}{2}\right) \), \( (1, 1) \)
\( x = 1 \Rightarrow y = 1 \)
(e) \( 4x^2 + 3xy + (2xy - 3x^2) i = 4y^2 - \frac{x^2}{2} + (3xy - 2y^2) i \)
\( 4x^2 + 3xy = 4y^2 - \frac{x^2}{2} \quad \& \quad 2xy - 3x^2 = 3xy - 2y^2 \)
By solving \( x = k \), \( y = \frac{3k}{2} \), \( k \in \mathbb{R} \)

Question. Show that all the roots of the equation
\( a_1z^3 + a_2z^2 + a_3z + a_4 = 3 \), where \( |a_i| \le 1, i = 1, 2, 3, 4 \)
lie outside the circle with centre origin and radius \( 2/3 \).
Answer: \( |a_1 z^3 + a_2z^2 + a_3z + a_4| = 3 \)
\( 3 \le |a_1| |z|^3 + |a_2| |z|^2 + |a_3| |z| + |a_4| \)
\( 3 < |z|^3 + |z|^2 + |z| + 1 \)
\( 3 < \frac{1 - |z|^4}{1 - |z|} \)
\( 1 - |z| > 0 \)
\( 1 - |z|^4 > 3 - 3|z| \)
\( 3|z| > 2 + |z|^4 \)
\( |z| > \frac{2}{3} + \frac{1}{3} |z|^4 \)
here \( |z| > \frac{2}{3} \)

Question. (a) Find all non-zero complex numbers Z satisfying \( \overline{Z} = iZ^2 \).
(b) If the complex numbers \( z_1, z_2, \dots, z_n \) lie on the unit circle \( |z| = 1 \) then show that \( |z_1 + z_2 + \dots + z_n| = |z_1^{-1} + z_2^{-1} + \dots + z_n^{-1}| \).
Answer: (a) \( \overline{z} = iz^2 \)
\( \Rightarrow |\overline{z}| = |iz^2| \Rightarrow |z| = |z|^2 \)
\( \Rightarrow r = r^2 \)
\( r = 1 \Rightarrow z\overline{z} = 1 \)
\( \overline{z} = 1/z \)
\( \frac{1}{z} = iz^2 \)
\( z^3 = -i \)
\( z = (-i)^{1/3} \Rightarrow z = i, \pm \frac{\sqrt{3}}{2} - \frac{i}{2} \)

(b) \( |z| = 1 \quad |z_1| = |z_2| = |z_3| = \dots = |z_n| = 1 \)
\( z_1 \overline{z}_1 = 1 \), \( z_2 \overline{z}_2 = 1 \dots z_n \overline{z}_n = 1 \)
\( |z_1^{-1} + z_2^{-1} + \dots + z_n^{-1}| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right| \)
\( = |\overline{z}_1 + \overline{z}_2 + \dots + \overline{z}_n| \)
\( = |\overline{z_1 + z_2 + \dots + z_n}| \)
\( = |z_1 + z_2 + \dots + z_n| = \text{LHS} \)

Question. Find the Cartesian equations of the locus of ‘z’ in the complex plane satisfying \( |z - 4| + |z + 4| = 16 \).
Answer: \( |z - 4| + |z + 4| = 16 \)
\( 2ae = 8 < 16 \) so locus is ellipse
\( 2a = 16 \Rightarrow a = 8 \)
\( 2ae = 8 \)
\( a^2e^2 = 16 \)
\( e^2 = \frac{1}{4} \)
\( 1 - \frac{b^2}{a^2} = \frac{1}{4} \Rightarrow b^2 = 48 \)
Equation of ellipse will be \( \frac{x^2}{64} + \frac{y^2}{48} = 1 \)

Question. If \( \omega \) is an imaginary cube root of unity then prove that
(a) \( (1 + \omega - \omega^2)^3 - (1 - \omega + \omega^2)^3 = 0 \)
(b) \( (1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 = 32 \)
(c) If \( \omega \) is the cube root of unity, Find the value of \( (1 + 5\omega^2 + \omega^4) (1 + 5\omega^4 + \omega^2) (5\omega^3 + \omega + \omega^2) \).
Answer: (a) \( (1 + \omega - \omega^2)^3 - (1 - \omega + \omega^2)^3 \)
\( = (-2\omega^2)^3 - (-2\omega)^3 \)
\( = -8 + 8 = 0 = \text{RHS} \)
(b) \( (1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 \)
\( = (-2\omega)^5 + (-2\omega^2)^5 \)
\( = (-2)^5 [\omega^2 + \omega] = 2^5 = 32 = \text{RHS} \)
(c) \( (1 + 5\omega^2 + \omega^4) (1 + 5\omega^4 + \omega^2) (5\omega^3 + \omega + \omega^2) \)
\( = (1 + \omega + \omega^2 + 4\omega^2) (1 + \omega + \omega^2 + 4\omega^4) (4 + 1 + \omega + \omega^2) \)
\( = (4\omega^2) (4\omega) (4) = 64 \)

Question. Locate the points representing the complex number z on the Argand plane
(a) \( |z + 1 - 2i| = \sqrt{7} \)
(b) \( |z - 1|^2 + |z + 1|^2 = 4 \)
(c) \( \left|\frac{z - 3}{z + 3}\right| = 3 \)
(d) \( |z - 3| = |z - 6| \)
Answer: (a) \( |z + 1 - 2i| = \sqrt{7} \)
It represent a circle with centre \( (-1, 2) \) & radius = \( \sqrt{7} \)
(b) \( |z - 1|^2 + |z + 1|^2 = 4 \)
\( (z - 1) (\overline{z} - 1) + (z + 1) (\overline{z} + 1) = 4 \)
\( z\overline{z} - z - \overline{z} + 1 + z\overline{z} + z + \overline{z} + 1 = 4 \)
\( 2z\overline{z} = 2 \)
\( z\overline{z} = 1 \)
\( x^2 + y^2 = 1 \)
circle with centre at origin and radius = 1
(c) \( \left|\frac{z - 3}{z + 3}\right| = 3 \)
\( |z - 3| = 3|z + 3| \)
\( |z - 3|^2 = 9|z + 3|^2 \)
\( (z - 3) (\overline{z} - 3) = 9 (z + 3) (\overline{z} + 3) \)
\( 8z\overline{z} + 30z + 30\overline{z} + 72 = 0 \)
\( 8(x^2 + y^2) + 30x + 72 = 0 \)
\( x^2 + y^2 + \frac{15}{2}x + 9 = 0 \)
centre = \( \left(-\frac{15}{4}, 0\right) \) Radius = \( \frac{9}{4} \)
(d) \( |z - 3| = |z - 6| \)
squaring both side
\( |z - 3|^2 = |z - 6|^2 \)
\( (z - 3) (\overline{z} - 3) = (z - 6) (\overline{z} - 6) \)
\( z\overline{z} - 3z - 3\overline{z} + 9 = z\overline{z} - 6z - 6\overline{z} + 36 \)
\( 3z + 3\overline{z} - 27 = 0 \)
\( z + \overline{z} - 9 = 0 \)
\( x + iy + x - iy - 9 = 0 \)
\( 2x - 9 = 0 \) a st. line.

Question. Find the modulus, argument and the principal argument of the complex numbers.
(i) \( 6(\cos 310^\circ - i \sin 310^\circ) \)
(ii) \( -2(\cos 30^\circ + i \sin 30^\circ) \)
(iii) \( \frac{2 + i}{4i + (1 + i)^2} \)
Answer: (i) \( 6(\cos 310^\circ - i \sin 310^\circ) \)
\( 6[\cos (2\pi - 50^\circ) - i \sin (2\pi - 50^\circ)] \)
\( 6[\cos 50^\circ + i \sin 50^\circ] \)
modulus = 6
principal Arg = \( \frac{5\pi}{18} \)
Arg = \( 2k\pi + \frac{5\pi}{18} \), where \( k \in \text{I} \)
(ii) \( -2(\cos 30^\circ + i \sin 30^\circ) \)
\( 2 \left[ -\cos \frac{\pi}{6} - i \sin \frac{\pi}{6} \right] \)
\( 2 \left[ \cos \left( \pi + \frac{\pi}{6} \right) + i \sin \left( \pi + \frac{\pi}{6} \right) \right] \)
\( 2 \left[ \cos \frac{7\pi}{6} + i \sin \frac{7\pi}{6} \right] \)
Modulus = 2
Principal Arg = \( \frac{7\pi}{6} \)
Arg = \( 2m\pi + \frac{7\pi}{6} \quad m \in \text{I} \)
(iii) \( z = \frac{2 + i}{4i + (1 + i)^2} \)
\( = \frac{2 + i}{4i + 1 - 1 + 2i} = \frac{2 + i}{6i} \)
\( = -\frac{i}{6} (2 + i) = -\frac{1}{6} (-1 + 2i) \)
Modulus = \( \sqrt{\frac{1}{36} + \frac{4}{36}} = \frac{\sqrt{5}}{6} \)
Principal Arg = \( -\tan^{-1} 2 \)
Arg z = \( 2k\pi - \tan^{-1} 2 \quad K \in \text{I} \)

Question. Prove that identity,
\( |1 - z_1 \overline{z}_2|^2 - |z_1 - z_2|^2 = (1 - |z_1|^2) (1 - |z_2|^2) \)
Answer: \( |1 - z_1 \overline{z}_2|^2 - |z_1 - z_2|^2 \)
\( = (1 - z_1 \overline{z}_2) (1 - \overline{z}_1 z_2) - (z_1 - z_2) (\overline{z}_1 - \overline{z}_2) \)
\( = 1 - \overline{z}_1 z_2 - z_1 \overline{z}_2 + |z_1|^2 |z_2|^2 - |z_1|^2 + z_1 \overline{z}_2 + \overline{z}_1 z_2 - |z_2|^2 \)
\( = 1 - |z_1|^2 + |z_1|^2 |z_2|^2 - |z_2|^2 \)
\( = (1 - |z_1|^2) (1 - |z_2|^2) = \text{RHS} \quad \text{H.P.} \)

Question. If \( \omega \) is a cube root of unity, prove that
(i) \( (1 + \omega - \omega^2)^3 - (1 - \omega + \omega^2)^3 = 0 \)
(ii) \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} = \omega^2 \)
(iii) \( (1 - \omega) (1 - \omega^2) (1 - \omega^4) (1 - \omega^8) = 9 \)
Answer: (i) \( (1 + \omega - \omega^2)^3 - (1 - \omega + \omega^2)^3 \)
\( = (-2\omega^2)^3 - (-2\omega)^3 = -8 + 8 = 0 \)
(ii) \( \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2} \)
\( = \frac{\omega^2(a + b\omega + c\omega^2)}{\omega^2(c + a\omega + b\omega^2)} = \omega^2 \)
(iii) \( (1 - \omega) (1 - \omega^2) (1 - \omega^4) (1 - \omega^8) \)
\( = (1 - \omega) (1 - \omega^2) (1 - \omega) (1 - \omega^2) \)
\( = (1 - \omega)^2 (1 - \omega^2)^2 \)
\( = (1 + \omega^2 - 2\omega) (1 + \omega^4 - 2\omega^2) \)
\( = (1 + \omega^2 - 2\omega) (1 + \omega - 2\omega^2) \)
\( = (-3\omega) (-3\omega^2) = 9 = \text{RHS} \)

Question. If \( x = a + b \); \( y = a\omega + b\omega^2 \); \( z = a\omega^2 + b\omega \), show that
(i) \( xyz = a^3 + b^3 \)
(ii) \( x^2 + y^2 + z^2 = 6ab \)
(iii) \( x^3 + y^3 + z^3 = 3 (a^3 + b^3) \)
Answer: \( z = a + b; y = a\omega + b\omega^2; z = a\omega^2 + b\omega \)
(i) \( xyz = (a + b) (a\omega + b\omega^2) (a\omega^2 + b\omega) \)
\( = (a + b) (a^2 + ab\omega^2 + ab\omega + b^2) \)
\( = (a + b) (a^2 - ab + b^2) = a^3 + b^3 \)
(ii) \( x^2 + y^2 + z^2 = (a + b)^2 + (a\omega + b\omega^2)^2 + (a\omega^2 + b\omega)^2 \)
\( = 6ab \)
(iii) \( x^3 + y^3 + z^3 = (a + b)^3 + (a\omega + b\omega^2)^3 + (a\omega^2 + b\omega)^3 \)
\( = (a + b)^3 + (a + b\omega)^3 + (a + b\omega^2)^3 \)
\( = 3a^3 + 3b^3 + 3ab (a + b) + 3 ab\omega (a + b\omega) + 3ab\omega^2 (a + b\omega^2) \)
\( = 3a^3 + 3b^3 = 3(a^3 + b^3) \)

Advanced Subjective Questions

Question. If \(\alpha\) & \(\beta\) are any two complex numbers, prove that
(i) \(|\alpha + \beta|^2 + |\alpha - \beta|^2 = 2(|\alpha|^2 + |\beta|^2)\)
(ii) \(\left| \alpha - \sqrt{\alpha^2 - \beta^2} \right| + \left| \alpha + \sqrt{\alpha^2 - \beta^2} \right| = |\alpha + \beta| + |\alpha - \beta|\).
Answer: (i) \(|\alpha + \beta|^2 + |\alpha - \beta|^2\)
\(= (\alpha + \beta) ( \overline{\alpha} + \overline{\beta} ) + (\alpha - \beta) ( \overline{\alpha} - \overline{\beta} )\)
\(= \alpha \overline{\alpha} + \beta \overline{\beta} + \alpha\overline{\beta} + \beta\overline{\alpha} + \alpha\overline{\alpha} + \beta\overline{\beta} - \alpha\overline{\beta} - \beta\overline{\alpha}\)
\(= 2(\alpha \overline{\alpha} + \beta \overline{\beta})\)
\(= 2(|\alpha|^2 + |\beta|^2)\)
(ii) Let \(z_1 = \alpha + \sqrt{\alpha^2 - \beta^2}\) : \(z_2 = \alpha - \sqrt{\alpha^2 - \beta^2}\)
\(z_1 + z_2 = 2\alpha\)
\(z_1 - z_2 = 2\sqrt{\alpha^2 - \beta^2}\)
\(z_1 z_2 = \beta^2\)
\(|z_1| + |z_2| = |\alpha + \beta| + |\alpha - \beta|\)
Squaring on both the side
\(|z_1|^2 + |z_2|^2 + 2 |z_1 z_2| = [|\alpha + \beta| + |\alpha - \beta|]^2\)
as we know that \(2 |z_1|^2 + 2|z_2|^2 = |z_1 + z_2|^2 + |z_1 - z_2|^2\)
LHS \(= \frac{1}{2} [|z_1 + z_2|^2 + |z_1 - z_2|^2 + 2|z_1z_2|]\)
\(= \frac{1}{2} [|2\alpha|^2 + 4 |\alpha^2 - \beta^2| + 2|\beta^2|]\)
\(= [2|\alpha|^2 + 2 |\beta|^2 + 2|\alpha + \beta| |\alpha - \beta|]\)
\(= |\alpha + \beta|^2 + |\alpha - \beta|^2 + 2|\alpha + \beta| |\alpha - \beta|\)
\(= [|\alpha + \beta| + |\alpha - \beta|]^2\) = RHS

Question. (a) \((1 + w)^7 = A + Bw\) where w is the imaginary cube root of of a unity and \(A, B \in \text{R}\), find the ordered pair \((A, B)\).
(b) The value of the expression ;
\(1 \cdot (2 - w) (2 - w^2) + 2 \cdot (3 - w) (3 - w^2) + \dots\dots + ( n - 1) \cdot (n - w) (n - w^2)\), where w is an imaginary cube root of unity is
Answer: (a) \((1 + w)^7 = A + B\omega\)
w is cube root of unity
so \(1 + w + w^2 = 0\)
\(1 + \omega = -\omega^2\)
\((-\omega^2)^7 = A + B\omega\)
\(-\omega^{14} = A + B\omega\)
\(-\omega^2 = A + B\omega\)
\((1 + \omega) = A + B\omega\)
ordered pair
\((A, B) = (1, 1)\)
(b) \(S = 1^3 - 1 + 2^3 - 1 + 3^3 - 1 \dots\dots + n^3 - 1\)
\(S = \left[ \frac{n(n + 1)}{2} \right]^2 - n\)

JEE Problems

Question. If \(i= \sqrt{-1}\), then \(4+5\left( -\frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{334} + 3\left( -\frac{1}{2} + \frac{i\sqrt{3}}{2} \right)^{365}\) is equal to
(a) \(1-i\sqrt{3}\)
(b) \(- 1 + i\sqrt{3}\)
(c) \(i\sqrt{3}\)
(d) \(- i\sqrt{3}\)
Answer: (c) \(i\sqrt{3}\)
Solution:
\(w = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = \omega\)
\(4 + 5 \omega^{334} + 3\omega^{365}\)
\(= 4 + 5\omega + 3\omega^2\)
\(= 1 + 3 + 3\omega + 3\omega^2 + 2\omega\)
\(= 1 + 2\omega\)
\(= 1 + (-1 + i\sqrt{3}) = i\sqrt{3}\)

Question. For complex numbers \(z\) & \(\omega\), prove that, \(|z|^2 \omega - |\omega|^2 z = z - \omega\) if and only if, \(z = \omega\) or \(z\bar{\omega} = 1\)
Answer:
\(|z|^2 = z \bar{z}\)
\(|z|^2 \omega - |\omega|^2 z = z - \omega\)
\(\Rightarrow z(1 + |\omega|^2) = \omega(1 + |z|^2)\)
\(\therefore \frac{z}{\omega} = \frac{1 + |z|^2}{1 + |\omega|^2} = \text{Real} \Rightarrow \frac{z}{\omega} = \overline{\left(\frac{z}{\omega}\right)} = \frac{\bar{z}}{\bar{\omega}}\)
\(z\bar{\omega} = \bar{z}\omega\) ...(1)
\(z(\bar{z}\bar{\omega} - 1) - \omega(\bar{\omega}\bar{z} - 1) = 0\)
\((z\bar{\omega} - 1)(z - \omega) = 0\) by Equation (1)
\(z = \omega \text{ or } z\bar{\omega} = 1\)

Question. If \(\alpha = e^{\frac{2\pi i}{7}}\) and \(f(x) = A_0 + \sum_{k=1}^{20} A_k x^k\), then find the value of, \(f(x) + f(\alpha x) +.......+ f(\alpha^6x)\) independent of \(\alpha\).
Answer:
\(\alpha = e^{\frac{i2\pi}{7}} \Rightarrow \alpha^7 = e^{i2\pi} = 1, \alpha \neq 1\)
or \(\alpha^7 - 1 = 0\) or \((\alpha - 1)(\alpha^6 + \alpha^5 + ... + 1) = 0\)
\(\Rightarrow 1 + \alpha + \alpha^2 + ... + \alpha^6 = 0\)
Also \(1 + \alpha^k + \alpha^{2k} + ... + \alpha^{6k} = \frac{1 - (\alpha^k)^7}{1 - \alpha^k} = \frac{1 - \alpha^{7k}}{1 - \alpha^k} = \frac{1 - 1}{1 - \alpha^k} = 0\) where \(k \neq 7m\)
\(f(x) = A_0 + \sum_{k=1}^{20} A_k x^k\)
Replace \(x\) by \(x, \alpha x, \alpha^2 x ... \alpha^6 x\) in the above and add then
L.H.S. \(= (A_0 + A_0 + ...) + \sum_{k=1}^{20} A_k x^k (1 + \alpha^k + \alpha^{2k} ... + \alpha^{6k})\)
\(= 7A_0 + 0 = 7A_0\)
Which is independent of \(\alpha\)

Question. Let \(\alpha + i\beta\) ; \(\alpha, \beta \in \mathbb{R}\), be a root of the equation \(x^3 + qx + r = 0\); \(q, r \in \mathbb{R}\). Find a real cubic equation, independent of \(\alpha\) & \(\beta\), whose one root is \(2\alpha\).
Answer:
Let roots \(\alpha + i\beta\), \(\alpha - i\beta\), \(\gamma\)
\(\alpha + i\beta + \alpha - i\beta + \gamma = 0\)
\(2\alpha + \gamma = 0 \Rightarrow \gamma = -2\alpha\)
again \((-2\alpha)^3 + q(-2\alpha) + r = 0\)
\(-(2\alpha)^3 - q(2\alpha) + r = 0 \Rightarrow (2\alpha)^3 + q(2\alpha) - r = 0\)
\(x^3 + qx - r = 0\)

Question. If \(z_1, z_2, z_3\) are complex number such that \(|z_1|=|z_2|=|z_3|= \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| =1\), then \(|z_1+z_2+z_3|\) is
(a) equal to 1
(b) less than 1
(c) greater than 3
(d) equal to 3
Answer: (a) equal to 1
Solution:
\(|z_1| = |z_2| = |z_3| = 1\)
\(\bar{z}_1 = \frac{1}{z_1} ; \bar{z}_2 = \frac{1}{z_2} ; \bar{z}_3 = \frac{1}{z_3}\)
\(\left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1\)
\(|\bar{z}_1 + \bar{z}_2 + \bar{z}_3| = 1 \Rightarrow |z_1 + z_2 + z_3| = 1\)

Question. If \(\text{arg}(z) < 0\) , then \(\text{arg}(- z) - \text{arg}(z)\) equals
(a) \(\pi\)
(b) \(-\pi\)
(c) \(-\frac{\pi}{2}\)
(d) \(\frac{\pi}{2}\)
Answer: (a) \(\pi\)
Solution:
\(\text{arg}(z) < 0 \Rightarrow z = x - iy\)
Let \(\text{arg}(z) = -\theta\), then \(-z = -x + iy\)
\(\text{arg}(-z) - \text{arg}(z) = (\pi - \theta) - (-\theta) = \pi\)

Question. Given, \(z = \cos\frac{2\pi}{2n+1} + i \sin\frac{2\pi}{2n+1}\), 'n' a positive integer, find the equation whose roots are, \(\alpha = z + z^3 + .... + z^{2n-1}\) & \(\beta = z^2 + z^4 +.......+ z^{2n}\).
Answer:
\(z = \cos\frac{2\pi}{2n+1} + i \sin\frac{2\pi}{2n+1}\)
Clearly \(z^{2n+1} = \cos 2\pi + i \sin 2\pi = 1\)
\(\alpha = \frac{z[1 - (z^2)^n]}{1 - z^2} = \frac{z - 1}{1 - z^2} = \frac{-1}{z+1}\)
\(\beta = \frac{z^2[1 - (z^2)^n]}{1 - z^2} = \frac{z - 1}{1 - z^2} \times (-z) = \frac{-z}{z+1}\)
\(\alpha + \beta = -\left(\frac{z+1}{z+1}\right) = -1\)
\(\alpha\beta = \frac{z}{(z+1)^2} = \frac{1}{z + \frac{1}{z} + 2}\)
\(z = \cos\theta + i \sin\theta\) where \(\theta = \frac{2\pi}{2n+1}\)
and \(\frac{1}{z} = \cos\theta - i\sin\theta \therefore z + \frac{1}{z} = 2\cos\theta\)
\(\alpha\beta = \frac{1}{2\cos\theta + 2} = \frac{1}{4\cos^2\frac{\theta}{2}}\)
Equation will be \(x^2 + x + \frac{1}{4\cos^2\frac{\pi}{2n+1}} = 0\)

Question. Find all those roots of the equation \(z^{12} - 56z^6 - 512 = 0\) whose imaginary part is positive.
Answer:
\(z^{12} - 56z^6 - 512 = 0\)
Let \(z^6 = t\)
\(t^2 - 56t - 512 = 0\)
\((t - 64)(t + 8) = 0\)
\(z^6 = 64 \Rightarrow z = 2(1)^{1/6}\)
\(z^6 = -8 \Rightarrow z = \sqrt{2}(-1)^{1/6}\)
\(z = 2[\cos 0 + i \sin 0]^{1/6} = 2 \left[ \cos\frac{2m\pi}{6} + i \sin\frac{2m\pi}{6} \right]\)
Where \(m = 0, 1, 2, 3, 4, 5\)
Img. part corresponding to \(m = 1, 2\) will be +ve. Hence we choose \(m = 1, 2\).
Again \(z = \sqrt{2}(-1)^{1/6} = \sqrt{2}[\cos\pi + i \sin\pi]^{1/6} = \sqrt{2} \left[ \cos\frac{(2m+1)\pi}{6} + i \sin\frac{(2m+1)\pi}{6} \right]\)
Img. part corresponding to \(m = 0, 1, 2\) will be +ve.

Question. Let \(\omega = - \frac{1}{2} + i \frac{\sqrt{3}}{2}\). Then the value of the determinant \(\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4 \end{vmatrix}\) is
(a) \(3\omega\)
(b) \(3\omega(\omega - 1)\)
(c) \(3\omega^2\)
(d) \(3\omega(1 - \omega)\)
Answer: (b) \(3\omega(\omega - 1)\)

Question. For all complex numbers \(z_1, z_2\) satisfying \(|z_1|=12\) and \(|z_2 - 3 - 4i| = 5\), the minimum value of \(|z_1 - z_2|\) is
(a) 0
(b) 2
(c) 7
(d) 17
Answer: (b) 2

Question. Let a complex number \(\alpha\), \(\alpha \neq 1\), be a root of the equation \(z^{p+q} - z^p - z^q + 1 = 0\), where p, q are distinct primes. Show that either \(1 + \alpha + \alpha^2 + ........ + \alpha^{p-1} = 0\) or \(1 + \alpha + \alpha^2 + ....... + \alpha^{q-1} = 0\), but not both together.
Answer: \(z^{p+q} - z^p - z^q + 1 = 0\)
\((z^p - 1) (z^q - 1) = 0\)
Either \(\alpha\) is a \(p^{\text{th}}\) root of unity or \(q^{\text{th}}\) roots of unity
using the prop. of \(n^{\text{th}}\) root of unity either \(1 + \alpha + \alpha^2 + ... + \alpha^{p-1} = 0\) or \(1 + \alpha + \alpha^2 + ... + \alpha^{q-1} = 0\)
If both the equation hold simultaneously, without loss of generalisation let \(p > q\)
\(\therefore 1 + \alpha + \alpha^2 + ... + \alpha^{p-1} = 0\)
\(\Rightarrow 1 + \alpha + \alpha^2 + ... + \alpha^{q-1} + \alpha^q + \alpha^{q+1} + ... + \alpha^{p-1} = 0\)
\(\Rightarrow 0 + \alpha^q + \alpha^{q+1} + ... + \alpha^{p-1} = 0\)
Now, \(\alpha^q = 1\)
\(\Rightarrow 1 + \alpha + \alpha^2 + ... + \alpha^{p-q-1} = 0\)
Hence \(\alpha\) should be the \((p - q)^{\text{th}}\) root of unity i.e. \(\alpha^{p-q} = 1\)
\(\Rightarrow p - q\) is a multiple of \(q\) (\(\because\) q is prime)
i.e. \(p - q = nq \Rightarrow p = (n + 1)q\)
\(\Rightarrow p\) is not prime which is a contradiction. Hence Proved.

Question. If \(z_1\) and \(z_2\) are two complex numbers such that \(|z_1| < 1 < |z_2|\) then prove that \(\left| \frac{1 - z_1 \bar{z}_2}{z_1 - z_2} \right| < 1\).
Answer: \(|z_1| < 1 < |z_2|\)
To Prove That: \(\left| \frac{1 - z_1 \bar{z}_2}{z_1 - z_2} \right| < 1\)
\(|1 - z_1 \bar{z}_2|^2 < |z_2 - z_1|^2\)
\((1 - z_1 \bar{z}_2) (1 - \bar{z}_1 z_2) < (z_2 - z_1)(\bar{z}_2 - \bar{z}_1)\)
\(1 + |z_1|^2 |z_2|^2 - z_1 \bar{z}_2 - \bar{z}_1 z_2 < |z_2|^2 + |z_1|^2 - z_2 \bar{z}_1 - z_1 \bar{z}_2\)
\(1 + |z_1|^2 |z_2|^2 - |z_2|^2 - |z_1|^2 < 0\)
\((1 - |z_1|^2) - |z_2|^2(1 - |z_1|^2) < 0\)
\((1 - |z_1|^2) (1 - |z_2|^2) < 0\)
Which is true because \(|z_1| < 1 < |z_2|\). Hence Proved.

Question. Prove that there exists no complex number z such that \(|z| < 1/3\) & \(\sum_{r=1}^n a_r z^r = 1\) where \(|a_r| < 2\).
Answer: \(1 = |\sum a_r z^r| \leq \sum |a_r| |z|^r\)
\(\Rightarrow 1 \leq |a_1| |z| + |a_2| |z|^2 + |a_3| |z|^3 + ...+ |a_n| |z|^n\)
\(< 2 (|z| + |z|^2 + |z|^3 + ... |z|^n)\)
\(1 < 2 \left( \frac{|z|}{1 - |z|} \right)\)
If \(|z| < 1/3 \Rightarrow 1 < 2 \left( \frac{1/3}{1 - 1/3} \right) \Rightarrow 1 < 1\) Which is not possible.
Hence \(|z| < 1/3\) and \(\sum a_r z^r = 1\) cannot occur simultaneously for any \(a_r\), \(|a_r| < 2\).

Question. \(\omega\) is an imaginary cube root of unity. If \((1 + \omega^2)^m = (1 + \omega^4)^m\), then the least positive integral value of m is
(a) 6
(b) 5
(c) 4
(d) 3
Answer: (d) 3 

Question. Find the centre and radius of circle determined by all complex numbers \(z = x + iy\) satisfying \(\left| \frac{z - \alpha}{z - \beta} \right| = k\), where \(\alpha = \alpha_1 + i\alpha_2\), \(\beta = \beta_1 + i\beta_2\) are fixed complex and \(k \neq 1\)
Answer: \(\left| \frac{z - \alpha}{z - \beta} \right| = k\)
\(|z - \alpha|^2 = k^2 |z - \beta|^2\)
\((z - \alpha)(\bar{z} - \bar{\alpha}) = k^2 (z - \beta) (\bar{z} - \bar{\beta})\)
\(z\bar{z} - \alpha\bar{z} - \bar{\alpha}z + \alpha\bar{\alpha} = k^2 (z\bar{z} - \beta\bar{z} - \bar{\beta}z + \beta\bar{\beta})\)
\(z\bar{z} - \left(\frac{\alpha - k^2\beta}{1 - k^2}\right)\bar{z} - \left(\frac{\bar{\alpha} - k^2\bar{\beta}}{1 - k^2}\right)z + \frac{\alpha\bar{\alpha} - k^2\beta\bar{\beta}}{1 - k^2} = 0\)
Which represents a circle with
centre \(z_0 = \frac{\alpha - k^2\beta}{1 - k^2}\)
and radius \(r = \sqrt{|z_0|^2 - \frac{\alpha\bar{\alpha} - k^2\beta\bar{\beta}}{1 - k^2}} = \sqrt{\frac{|\alpha - k^2\beta|^2}{(1 - k^2)^2} - \frac{\alpha\bar{\alpha} - k^2\beta\bar{\beta}}{1 - k^2}} = \left| \frac{k(\alpha - \beta)}{1 - k^2} \right|\)

Question. If a, b, c are integers not all equal and \(\omega\) is cube root of unity (\(\omega \neq 1\)), then the minimum value of \(|a + b\omega + c\omega^2|\) is
(a) 0
(b) 1
(c) \(\sqrt{3}/2\)
(d) 1/2
Answer: (b) 1

Question. If one of the vertices of the square circumscribing the circle \(|z - 1| = \sqrt{2}\) is \(2 + \sqrt{3} i\). Find the other vertices of the square.
Answer: Since centre of circle i.e. (1, 0) is also the mid point of diagonals of square
\(\Rightarrow \frac{z_1 + z_2}{2} = z_0 \Rightarrow z_2 = -\sqrt{3}i\)
and \(\frac{z_3 - 1}{z_1 - 1} = e^{\pm i\frac{\pi}{2}}\)
\(\Rightarrow\) other vertices are \(z_3, z_4 = (1 - \sqrt{3}) + i\) and \((1 + \sqrt{3}) - i\)

Question. If \(w = \alpha + i\beta\) where \(\beta \neq 0\) and \(z \neq 1\), satisfies the condition that \(\frac{w - \bar{w}z}{1 - z}\) is purely real , then the set of the values of z is
(a) \(\{z : |z| = 1\}\)
(b) \(\{z : z = \bar{z}\}\)
(c) \(\{z : z \neq 1\}\)
(d) \(\{z : |z| = 1, z \neq 1\}\)
Answer: (d) \(\{z : |z| = 1, z \neq 1\}\)

Question. A man walks a distance of 3 units from the origin towards the north-east (N 45° E) direction. Form there, he walks a distance of 4 units towards the north-west (N 45° W) direction to reach a point P. Then the position of P in the Argand plane is
(a) \(3e^{i\pi/4} + 4i\)
(b) \((3 - 4i)e^{i\pi/4}\)
(c) \((4 + 3i)e^{i\pi/4}\)
(d) \((3 + 4i)e^{i\pi/4}\)
Answer: (d) \((3 + 4i)e^{i\pi/4}\)

Question. If \(|z| = 1\) and \(z \neq \pm 1\), then all the values of \(\frac{z}{1 - z^2}\) lie on
(a) a line not passing through the origin
(b) \(|z| = \sqrt{2}\)
(c) the x-axis
(d) the y-axis
Answer: (d) the y-axis

Question. A particle P starts from the point \(z_0 = 1 + 2i\), where \(i = \sqrt{-1}\). It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point \(z_1\). From \(z_1\) the particle moves \(\sqrt{2}\) units in the direction of the vector \(\hat{i} + \hat{j}\) and then it moves through an angle \(\frac{\pi}{2}\) in anticlockwise direction on a circle with centre at origin, to reach a point \(z_2\). The point \(z_2\) is given by
(a) \(6 + 7i\)
(b) \(-7 + 6i\)
(c) \(7 + 6i\)
(d) \(-6 + 7i\)
Answer: (d) \(-6 + 7i\)

Comprehension (3 questions together)
Let A, B, C be three sets of complex numbers as defined below
A = \(\{z : \text{Im } z \geq 1\}\)
B = \(\{z : |z - 2 - i| = 3\}\)
C = \(\{z : \text{Re}((1 - i)z) = \sqrt{2}\}\).

Question. The number of elements in the set \(A \cap B \cap C\) is
(a) 0
(b) 1
(c) 2
(d) \(\infty\)
Answer: (b) 1
Solution:
A = Set of points on and above the line \(y = 1\)
B = set of points on the circle \((x - 2)^2 + (y - 1)^2 = 9\)
C : \(\text{Re}((1 - i)(x + iy)) = x + y = \sqrt{2}\)
Intersection of \(A \cap B \cap C\) has only one point of intersection.

Question. Let z be any point in \(A \cap B \cap C\). Then \(|z + 1 - i|^2 + |z - 5 - i|^2\) lies between
(a) 25 & 29
(b) 30 & 34
(c) 35 & 39
(d) 40 & 44
Answer: (c) 35 & 39
Solution:
The points \((-1, 1)\) and \((5, 1)\) are the extremities of a diameter of the given circle.
Hence \(|z + 1 - i|^2 + |z - 5 - i|^2 = 36\). This lies between 35 and 39.

Question. Let z be any point in \(A \cap B \cap C\) and let w be any point satisfying \(|w - 2 - i| < 3\). Then, \(|z| - |w| + 3\) lies between
(a) -6 & 3
(b) -3 & 6
(c) -6 & 6
(d) -3 & 9
Answer: (d) -3 & 9
Solution:
\(||z| - |w|| < |z - w|\)
and \(|z - w|\) = Distance between z and w.
z is fixed. Hence distance between z and w would be maximum for diametrically opposite point.
\(\Rightarrow |z - w| < 6 \Rightarrow -6 < |z| - |w| < 6\)
\(-3 < |z| - |w| + 3 < 9\)