# Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:

Question:

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:

(i) $1+i$

(ii) $\sqrt{3}+i$

(iii) $1-i$

(iv) $\frac{1-i}{1+i}$

(v) $\frac{1}{1+i}$

(vi) $\frac{1+2 i}{1-3 i}$

(vii) $\sin 120^{\circ}-i \cos 120^{\circ}$

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

Solution:

(i) $z=1+i$

$r=|z|$

$=\sqrt{1+1}$

$=\sqrt{2}$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\Rightarrow \tan \alpha=\left(\frac{1}{1}\right)$

$\Rightarrow \alpha=\frac{\pi}{4}$

Since point $(1,1)$ lies in the first quadrant, the argument of $z$ is given by $\theta=\alpha=\frac{\pi}{4}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$

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

$r=|z|$

$=\sqrt{3+1}$

$=\sqrt{4}$

$=2$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\Rightarrow \tan \alpha=\left(\frac{1}{\sqrt{3}}\right)$

$\Rightarrow \alpha=\frac{\pi}{6}$

Since point $(\sqrt{3}, 1)$ lies in the first quadrant, the argument of $z$ is given by $\theta=\alpha=\frac{\pi}{6}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)$

(iii) $z=1-i$

$r=|z|$

$=\sqrt{1+1}$

$=\sqrt{2}$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\therefore \tan \alpha=\left|\frac{-1}{1}\right|$

$=\frac{\pi}{4}$

$\Rightarrow \alpha=\frac{\pi}{4}$

Since point $(1,-1)$ lies in the fourth quadrant, the argument of $z$ is given by $\theta=-\alpha=-\frac{\pi}{4}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=\sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right\}$

$=\sqrt{2}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)$

(iv) $\frac{1-i}{1+i}$

Rationalising the denominator :

$\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$\Rightarrow \frac{1+i^{2}-2 i}{1-i^{2}}$

$\Rightarrow \frac{-2 i}{2} \quad\left(\because i^{2}=-1\right)$

$\Rightarrow-i$

$r=|z|$

$=\sqrt{0+1}$

$=1$

Since point $(0,-1)$ lies on the negative direction of the imaginary axis, the argument of $z$ is given by $\frac{3 \pi}{2}$.

Polar form $=r(\cos \theta+i \sin \theta)$

$=\left(c \cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)$

$=\left\{c \operatorname{os}\left(2 \pi-\frac{\pi}{2}\right)+i \sin \left(2 \pi-\frac{\pi}{2}\right)\right\}$

$=\left(\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}\right)$

$(v) \frac{1}{1+i}$

Rationalising the denominator:

$\frac{1}{1+i} \times \frac{1-i}{1-i}$

$\Rightarrow \frac{1-i}{1-i^{2}}$

$\Rightarrow \frac{1-i}{2} \quad\left(\because i^{2}=-1\right)$

$\Rightarrow \frac{1}{2}-\frac{i}{2}$

$r=|z|$

$=\sqrt{\frac{1}{4}+\frac{1}{4}}$

$=\frac{1}{\sqrt{2}}$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\therefore \tan \alpha=\left|\frac{\frac{1}{2}}{\frac{-1}{2}}\right|$

= 1

$\Rightarrow \alpha=\frac{\pi}{4}$

Since point $\left(\frac{1}{2},-\frac{1}{2}\right)$ lies in the fourth quadrant, the argument is given by $\theta=-\alpha=\frac{-\pi}{4}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=\frac{1}{\sqrt{2}}\left\{c \cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right\}$

$=\frac{1}{\sqrt{2}}\left(c \cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)$

$(v i) \frac{1+2 i}{1-3 i}$

Rationalising the denominator:

$\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}$

$\Rightarrow \frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}}$

$\Rightarrow \frac{-5+5 i}{10} \quad\left(\because i^{2}=-1\right)$

$\Rightarrow \frac{-1}{2}+\frac{i}{2}$

$r=|z|$

$=\sqrt{\frac{1}{4}+\frac{1}{4}}$

$=\frac{1}{\sqrt{2}}$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

Then, $\tan \alpha=\left|\frac{\frac{1}{2}}{\frac{-1}{2}}\right|$

= 1

$\Rightarrow \alpha=\frac{\pi}{4}$

Since point $\left(\frac{-1}{2}, \frac{1}{2}\right)$ lies in the second quadrant, the argument is given by

$\theta=\pi-\alpha$

$=\pi-\frac{\pi}{4}$

$=\frac{3 \pi}{4}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=\frac{1}{\sqrt{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

$($ vii $) \sin 120^{\circ}-i \cos 120^{\circ}$

$\frac{\sqrt{3}}{2}+\frac{i}{2}$

$r=|z|$

$=\sqrt{\frac{3}{4}+\frac{1}{4}}$

$=1$

Let $\tan \alpha=\left|\frac{\operatorname{lm}(z)}{\operatorname{Re}(z)}\right|$

Then, $\tan \alpha=\left|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|$

$=\frac{1}{\sqrt{3}}$

$\Rightarrow \alpha=\frac{\pi}{6}$

Since point $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ lies in the first quadrant, the argument is given by $\theta=\alpha=\frac{\pi}{6}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}$

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

Rationalising the denominator:

$\frac{-16}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$

$\Rightarrow \frac{-16+16 \sqrt{3} i}{1-3 i^{2}}$

$\Rightarrow \frac{-16+16 \sqrt{3} i}{4}$    $\left(\because i^{2}=-1\right)$

$\Rightarrow-4+4 \sqrt{3} i$

$r=|z|$

$=\sqrt{16+48}$

$=8$

Let $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

Then, $\tan \alpha=\left|\frac{4 \sqrt{3}}{-4}\right|$

$=\sqrt{3}$

$\Rightarrow \alpha=\frac{\pi}{3}$

Since the point $(-4,4 \sqrt{3})$ lies in the third quadrant, the argument is given by $\theta=\pi-\alpha$

$=\pi-\frac{\pi}{3}$

$=\frac{2 \pi}{3}$

Polar form $=r(\cos \theta+i \sin \theta)$

$=8\left\{c \cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right\}$