Polar representation of a complex number | Part-3

Converting Complex Numbers to Polar Form

(i) \( \frac{-16}{1 + i\sqrt{3}} \)

First, multiply numerator and denominator by the conjugate of the denominator:
\[ \frac{-16}{1 + i\sqrt{3}} \cdot \frac{1 - i\sqrt{3}}{1 - i\sqrt{3}} = \frac{-16(1 - i\sqrt{3})}{1^2 - (i\sqrt{3})^2} = \frac{-16(1 - i\sqrt{3})}{1 + 3} \]
\[ = \frac{-16(1 - i\sqrt{3})}{4} = -4(1 - i\sqrt{3}) = -4 + 4i\sqrt{3} \]

Let \( x = -4, y = 4\sqrt{3} \)
\( r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \)
\( \cos\theta = \frac{-4}{8} = -\frac{1}{2},\ \sin\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2} \)
So, \( \theta = \frac{2\pi}{3} \)

Polar Form: \( z = 8\left( \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right) \)

(ii) \( \frac{1 + 7i}{(2 - i)^2} \)

First compute \( (2 - i)^2 \):
\[ (2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i \] So we compute:
\[ \frac{1 + 7i}{3 - 4i} \] Multiply numerator and denominator by the conjugate:
\[ \frac{1 + 7i}{3 - 4i} \cdot \frac{3 + 4i}{3 + 4i} = \frac{(1 + 7i)(3 + 4i)}{3^2 + 4^2} \] Numerator:
\[ 1 \cdot 3 + 1 \cdot 4i + 7i \cdot 3 + 7i \cdot 4i = 3 + 4i + 21i + 28i^2 = 3 + 25i - 28 = -25 + 25i \] Denominator: \( 9 + 16 = 25 \)
So: \[ \frac{-25 + 25i}{25} = -1 + i \]

Let \( x = -1, y = 1 \)
\( r = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)
\( \cos\theta = -\frac{1}{\sqrt{2}},\ \sin\theta = \frac{1}{\sqrt{2}} \Rightarrow \theta = \frac{3\pi}{4} \)

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

(iii) \( \frac{i - 1}{\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}} \)

Use the identity: \( \cos\theta + i\sin\theta = \text{cis}(\theta) \)
So this is: \[ \frac{i - 1}{\text{cis}\left(\frac{\pi}{3}\right)} \] Let’s write numerator in rectangular form: \( -1 + i \)
First, write \( -1 + i \) in polar form:
\( r = \sqrt{1^2 + 1^2} = \sqrt{2} \), and since it lies in 2nd quadrant: \( \theta_1 = \frac{3\pi}{4} \)
So: \[ \frac{-1 + i}{\text{cis}\left(\frac{\pi}{3}\right)} = \frac{\sqrt{2}\ \text{cis}\left(\frac{3\pi}{4}\right)}{\text{cis}\left(\frac{\pi}{3}\right)} = \sqrt{2}\ \text{cis}\left(\frac{3\pi}{4} - \frac{\pi}{3}\right) \]
\[ = \sqrt{2}\ \text{cis}\left(\frac{5\pi}{12}\right) \]

Polar Form: \( z = \sqrt{2}\left( \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} \right) \)