Question

If \( z_1 = 2\sqrt{2}(1 + i) \) and \( z_2 = 1 + i\sqrt{3} \), then \( z_1^2 z_2^3 \) is equal to:

• A. \( 128i \)
• B. \( 64i \)
• C. \( -64i \)
• D. \( -128i \)
• E. \( 256 \)

Hint:

Remember that $e^{i\pi/2} = i$ and $e^{i\pi} = -1$. These rotations on the unit circle allow you to switch back to rectangular form instantly.

Answer 1

The Correct Option is D

Understanding the Concept: Complex numbers are often easiest to manipulate in polar or exponential form \( r e^{i\theta} \) when dealing with powers and products. For \( z = x + iy \), the magnitude is \( r = \sqrt{x^2 + y^2} \) and the argument is \( \theta = \tan^{-1}(y/x) \).

Step 1: Convert \( z_1 \) and \( z_2 \) to polar form.
For \( z_1 = 2\sqrt{2} + i2\sqrt{2} \): \[ r_1 = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8 + 8} = 4 \] \[ \theta_1 = \tan^{-1}(1) = \pi/4 \quad \Rightarrow \quad z_1 = 4e^{i\pi/4} \] For \( z_2 = 1 + i\sqrt{3} \): \[ r_2 = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \] \[ \theta_2 = \tan^{-1}(\sqrt{3}) = \pi/3 \quad \Rightarrow \quad z_2 = 2e^{i\pi/3} \]

Step 2: Calculate the powers.
\[ z_1^2 = (4e^{i\pi/4})^2 = 16e^{i\pi/2} = 16i \] \[ z_2^3 = (2e^{i\pi/3})^3 = 8e^{i\pi} = -8 \]

Step 3: Multiply the results.
\[ z_1^2 z_2^3 = (16i)(-8) = -128i \]