Question

If \( z(2 - i) = (3 + i) \), then \( z^{38} = \), (where \( z = x + iy \))

• A. \( - (2^{19})i \)
• B. \( 2^{19}i \)
• C. \( - (2^{19}) \)
• D. \( 2^{19} \)

Hint:

When raising a complex number to a high power, convert to polar form. Remember \(e^{i(9\pi + \pi/2)} = e^{i9\pi} \cdot e^{i\pi/2} = (-1) \cdot i = -i\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question: We solve for \(z\) from the given equation and then compute \(z^{38}\).

Step 2: Key Formula or Approach: First find \(z = \frac{3+i}{2-i}\) by multiplying numerator and denominator by the complex conjugate. Then write \(z\) in polar form \(re^{i\theta}\) to raise it to a power.

Step 3: Detailed Explanation: \[ z = \frac{3+i}{2-i} \cdot \frac{2+i}{2+i} = \frac{(3+i)(2+i)}{(2-i)(2+i)} = \frac{6+3i+2i+i^2}{4+1} = \frac{6+5i-1}{5} = \frac{5+5i}{5} = 1+i. \] Modulus: \(|z| = \sqrt{1^2+1^2} = \sqrt{2}\). Argument: \(\theta = \frac{\pi}{4}\) (since both real and imaginary parts are positive). Thus \(z = \sqrt{2} e^{i\pi/4}\). Now \(z^{38} = (\sqrt{2})^{38} e^{i 38\pi/4} = 2^{19} e^{i 19\pi/2}\). Simplify the angle: \(\frac{19\pi}{2} = 9\pi + \frac{\pi}{2}\). \(e^{i9\pi} = e^{i(8\pi+\pi)} = e^{i\pi} = -1\). And \(e^{i\pi/2} = i\). Hence \(e^{i 19\pi/2} = (-1) \cdot i = -i\). Therefore \(z^{38} = 2^{19} \cdot (-i) = - (2^{19}) i\).

Step 4: Final Answer: Option (A) is correct.