Question

Polar form of the complex number \(z = -4 + i4\sqrt{3}\) is :

• A. \(8(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})\)
• B. \(8(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})\)
• C. \(8(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3})\)
• D. \(8(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6})\)

Hint:

To save time, check the signs of \(x\) and \(y\) first:
\(x = -4\) (negative) and \(y = 4\sqrt{3}\) (positive) puts the point in the second quadrant.
The only second-quadrant angle among the choices is \(\frac{2\pi}{3}\) (since \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\) are in Quadrant I, and \(\frac{5\pi}{6}\) would require different coordinates).
This immediately points to Option (C).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To convert a complex number from its rectangular form \(z = x + iy\) to its polar form \(z = r(\cos\theta + i\sin\theta)\), we must find:
1. The modulus \(r = |z| = \sqrt{x^2 + y^2}\).
2. The argument \(\theta = \text{arg}(z)\), taking into account the quadrant in which the complex number lies.
Key Formula or Approach:
For \(z = x + iy\):
\[ r = \sqrt{x^2 + y^2} \]
The reference angle \(\alpha\) is given by:
\[ \alpha = \tan^{-1}\left|\frac{y}{x}\right| \]
Depending on the sign of \(x\) and \(y\), we find \(\theta\):
- Quadrant I (\(x > 0, y > 0\)): \(\theta = \alpha\)
- Quadrant II (\(x 0\)): \(\theta = \pi - \alpha\)
- Quadrant III (\(x < 0, y < 0\)): \(\theta = -\pi + \alpha\)
- Quadrant IV (\(x > 0, y < 0\)): \(\theta = -\alpha\)

Step 2: Detailed Explanation:
Let us apply this approach to \(z = -4 + i4\sqrt{3}\).
Here, \(x = -4\) and \(y = 4\sqrt{3}\).
Calculate the modulus \(r\):
\[ r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + (16 \times 3)} = \sqrt{16 + 48} = \sqrt{64} = 8 \]
Determine the quadrant:
Since \(x = -4 < 0\) and \(y = 4\sqrt{3} > 0\), the complex number lies in the second quadrant (Quadrant II).
Calculate the reference angle \(\alpha\):
\[ \alpha = \tan^{-1}\left|\frac{4\sqrt{3}}{-4}\right| = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Calculate the argument \(\theta\) for Quadrant II:
\[ \theta = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]
Now, substitute \(r\) and \(\theta\) into the polar form:
\[ z = 8\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) \]
This matches Option (C).

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