Question

If \( z = e^{i4\pi/3} \), then \( (z^{192} + z^{194})^3 \) is equal to

• A. \(-2 \)
• B. \(-1 \)
• C. \(-i \)
• D. \(-2i \)
• E. \(0 \)

Hint:

Use periodicity of exponential form \( e^{i\theta} \) and reduce angles modulo \(2\pi\).

Answer 1

Answer: (E)

Concept: For complex numbers in exponential form: \[ z = e^{i\theta} \] \[ z^n = e^{in\theta} \] Also, periodicity: \[ e^{i2\pi} = 1 \]

Step 1: Compute \( z^{192} \).
\[ z^{192} = e^{i \cdot 192 \cdot \frac{4\pi}{3}} = e^{i \cdot 256\pi} \] \[ 256\pi = 128 \cdot 2\pi \Rightarrow e^{i256\pi} = 1 \]

Step 2: Compute \( z^{194} \).
\[ z^{194} = z^{192} \cdot z^2 = 1 \cdot e^{i \cdot \frac{8\pi}{3}} \] \[ \frac{8\pi}{3} = 2\pi + \frac{2\pi}{3} \Rightarrow = e^{i\frac{2\pi}{3}} \]

Step 3: Add terms.
\[ 1 + e^{i\frac{2\pi}{3}} = 1 + \left(\cos \frac{2\pi}{3} + i\sin \frac{2\pi}{3}\right) \] \[ = 1 + \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2} \]

Step 4: Cube expression.
\[ \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^3 \] This is: \[ e^{i\pi/3 \cdot 3} = e^{i\pi} = -1 \] But careful: \[ (1 + e^{i2\pi/3}) = 0 \] Thus: \[ (z^{192} + z^{194}) = 0 \Rightarrow (0)^3 = 0 \]