Polar representation of a complex number | Part-2

Converting Complex Numbers into Polar Form

(i) \( z = \sqrt{3} + i \)

Let \( x = \sqrt{3},\ y = 1 \)
So, \( r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \)
Now, \( \cos\theta = \frac{x}{r} = \frac{\sqrt{3}}{2} \), and \( \sin\theta = \frac{y}{r} = \frac{1}{2} \)
Therefore, \( \theta = \frac{\pi}{6} \)

Polar Form: \( z = 2\left( \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} \right) \)

(ii) \( z = -\sqrt{3} + i \)

Let \( x = -\sqrt{3},\ y = 1 \)
\( r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \)
\( \cos\theta = \frac{-\sqrt{3}}{2},\ \sin\theta = \frac{1}{2} \)
This lies in the 2nd quadrant, so \( \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Polar Form: \( z = 2\left( \cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6} \right) \)

(iii) \( z = -1 - i\sqrt{3} \)

Let \( x = -1,\ y = -\sqrt{3} \)
\( r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \)
\( \cos\theta = \frac{-1}{2},\ \sin\theta = \frac{-\sqrt{3}}{2} \)
This lies in the 3rd quadrant, so \( \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \)

Polar Form: \( z = 2\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right) \)

(iv) \( z = -3 \)

Here, \( x = -3,\ y = 0 \)
\( r = \sqrt{(-3)^2 + 0^2} = \sqrt{9} = 3 \)
\( \cos\theta = \frac{-3}{3} = -1,\ \sin\theta = 0 \)
So, \( \theta = \pi \)

Polar Form: \( z = 3\left( \cos\pi + i\sin\pi \right) \)