Question
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: $\frac{-16}{1+\text{i}\sqrt{3}}$

Answer

The polar form of a complex number $\text{z}=\text{x}+\text{iy},$ is give by $\text{z}=|\text{z}|\big(\cos\theta+\text{i}\sin\theta\big)$ Where, $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$ and $\text{arg(z)}=\theta=\tan^{-1}\Big(\frac{\text{b}}{\text{a}}\Big)$ Let $\text{z}=\frac{-16}{1+\text{i}\sqrt{3}}$ $=\frac{-16}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$ $=\frac{-16(1-\text{i}\sqrt{3})}{(1)^2+(\sqrt{3})^2}$ $=\frac{-16(1-\text{i}\sqrt{3})}{1+3}$ $=\frac{-16}{4}(1-\text{i}\sqrt{3})$ $=-4(1-\text{i}\sqrt{3})$ $=-4+4\sqrt{3}\text{i}$ $\therefore \ |\text{z}|=\sqrt{(-4)^2+(4\sqrt{3})^2}$ $=\sqrt{16+48}$ $=\sqrt{64}$ $=8$ Here $\text{x}=-4<0$ & $\text{y}=4\text{x}3>0, \ \therefore\theta$ lies in quadrantII $=\theta=\text{arg(z)}=\tan^{-1}\Big(\frac{4\sqrt{3}}{-4}\Big)$ $=\tan^{-1}(\sqrt{3})$ $=\tan^{-1}\Big(\tan\big(\pi-\frac{\pi}{3}\big)\Big) \ \big(\therefore\tan(\pi-\theta)=-\tan\theta\big)$ $=\pi-\frac{\pi}{3}$ $=\frac{2\pi}{3}$ The polar form is given by $\text{z}=8\Big(\cos\frac{2\pi}{3}+\text{i}\sin\frac{2\pi}{3}\Big)$

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