Question

Match List-I with List-II:

\[ \begin{array}{|l|l|} \hline \textbf{List-I (Equation)} & \textbf{List-II (Roots)} \\ \hline (A)\; (z + 1)^{2n} + (z - 1)^{2n} = 0 & (I)\; \cos\!\left(\tfrac{\pi}{5}\right) \pm i\sin\!\left(\tfrac{\pi}{5}\right), -1, \cos\!\left(\tfrac{3\pi}{5}\right) \pm i\sin\!\left(\tfrac{3\pi}{5}\right) \\ \hline (B)\; z^5 + z^4 + z^3 + z^2 + z + 1 = 0 & (II)\; \text{purely imaginary number} \\ \hline (C)\; (z-1)^5 + z^5 = 0 & (III)\; -1,\; \pm\frac{1}{2} \pm i\frac{\sqrt{3}}{2} \\ \hline (D)\; z^5 + 1 = 0 & (IV)\; \tfrac{1}{2}\left[1+i\cot\!\left(\tfrac{p\pi}{10}\right)\right], \; p = 1,3,5,7,9 \\ \hline \end{array} \]
Choose the correct answer from the options given below:

• (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
• (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
• (A) - (IV), (B) - (II), (C) - (III), (D) - (I)
• (A) - (I), (B) - (III), (C) - (IV), (D) - (II)

Hint:

Problems involving sums or differences of powers of complex binomials, like \((z+a)^n \pm (z-a)^n = 0\), are often simplified by rearranging to \((\frac{z+a}{z-a})^n = \mp 1\) and then solving for z.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires solving four different equations in the complex domain and matching them to the correct description or set of roots.

Step 2: Detailed Explanation:
(A) \((z + 1)^{2n} + (z - 1)^{2n} = 0\):
This implies \(\left(\frac{z+1}{z-1}\right)^{2n} = -1 = e^{i(2k+1)\pi}\). So, \(\frac{z+1}{z-1} = e^{i\frac{(2k+1)\pi}{2n}}\). Let this ratio be \(w\). Since \(|w|=1\), solving gives \[ z = \frac{w+1}{w-1}. \] If \(w = \cos\theta + i\sin\theta\), then simplifying leads to \[ z = -i \cot\left(\tfrac{\theta}{2}\right). \] The roots are purely imaginary. (A) matches (II).

(B) \( z^5 + z^4 + z^3 + z^2 + z + 1 = 0 \):
This is the sum of a geometric series: \[ \frac{z^6-1}{z-1} = 0 \implies z^6=1, \; z\neq 1. \] Thus, roots are the 6th roots of unity excluding 1. \[ z = e^{i2\pi k/6}, \; k=1,2,3,4,5. \] Explicitly: \[ e^{i\pi/3}, \; e^{i2\pi/3}, \; e^{i\pi}=-1, \; e^{i4\pi/3}, \; e^{i5\pi/3}. \] In Cartesian form: \(-1, \pm\tfrac{1}{2} \pm i\tfrac{\sqrt{3}}{2}\). (B) matches (III).

(C) \( (z-1)^5 + z^5 = 0 \):
This gives \[ \left(\frac{z-1}{z}\right)^5 = -1 = e^{i(2k+1)\pi}. \] So, \[ 1 - \tfrac{1}{z} = e^{i\frac{(2k+1)\pi}{5}}, \quad \frac{1}{z} = 1 - w_k, \quad z = \frac{1}{1-w_k}. \] Let \(\theta_k = \tfrac{(2k+1)\pi}{5}\). Simplification gives: \[ z = \tfrac{1}{2}\left(1+i\cot\left(\tfrac{\theta_k}{2}\right)\right). \] Thus, \[ z = \tfrac{1}{2}\left(1+i\cot\left(\tfrac{p\pi}{10}\right)\right), \; p=1,3,5,7,9. \] (C) matches (IV).

(D) \( z^5 + 1 = 0 \):
\[ z^5 = -1 = e^{i\pi}. \] So, \[ z = e^{i\frac{(2k+1)\pi}{5}}, \; k=0,1,2,3,4. \] The roots are \[ e^{i\pi/5}, \; e^{i3\pi/5}, \; e^{i\pi}=-1, \; e^{i7\pi/5}, \; e^{i9\pi/5}. \] These correspond to \(-1, \cos(\pi/5)\pm i\sin(\pi/5), \cos(3\pi/5)\pm i\sin(3\pi/5)\). (D) matches (I).

Step 3: Final Answer:
The correct matching is: \[ (A)-(II), \quad (B)-(III), \quad (C)-(IV), \quad (D)-(I). \] Correct Choice: Option (A).