Question

The modulus and argument of complex number $z = \frac{1}{(2\sqrt{3} + 2i)^4}$ is}

• A. $|z| = \frac{1}{2^{16}}$, $\text{Arg}(z) = \frac{\pi}{3}$
• B. $|z| = \frac{1}{2^8}$, $\text{Arg}(z) = \frac{5\pi}{12}$
• C. $|z| = \frac{1}{2^8}$, $\text{Arg}(z) = \frac{4\pi}{3}$
• D. $|z| = \frac{1}{2^8}$, $\text{Arg}(z) = \frac{2\pi}{3}$

Hint:

Using polar coordinates and De Moivre's Theorem is much simpler and less error-prone than algebraically expanding high-power binomials.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the modulus and argument of a complex quotient or power, we can represent the base complex number in polar form first.
Key Formula or Approach:
For any complex number $w = x + iy$:
\[ |w| = \sqrt{x^2 + y^2} \]
\[ \text{Arg}(w) = \tan^{-1}\left(\frac{y}{x}\right) \]
For $z = w^{-4}$:
\[ |z| = |w|^{-4} \]
\[ \text{Arg}(z) = -4 \cdot \text{Arg}(w) \pmod{2\pi} \]

Step 2: Detailed Explanation:
Let the base complex number be $w = 2\sqrt{3} + 2i$.
1. Calculate the modulus of $w$:
\[ |w| = \sqrt{(2\sqrt{3})^2 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4 \]
2. Calculate the argument of $w$:
Since $w$ lies in the first quadrant:
\[ \theta = \tan^{-1}\left(\frac{2}{2\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Now, let us calculate the modulus of $z = w^{-4}$:
\[ |z| = |w|^{-4} = 4^{-4} = (2^2)^{-4} = 2^{-8} = \frac{1}{2^8} \]
Calculate the argument of $z$:
\[ \text{Arg}(z) = -4 \cdot \theta = -4 \cdot \frac{\pi}{6} = -\frac{2\pi}{3} \]
Convert this negative angle into its principal value within the range $[0, 2\pi)$:
\[ \text{Arg}(z) = 2\pi - \frac{2\pi}{3} = \frac{4\pi}{3} \]
Thus, $|z| = \frac{1}{2^8}$ and $\text{Arg}(z) = \frac{4\pi}{3}$.

Step 3: Final Answer
The correct option is (C).