Question

Find the modulus and argument of mentioned below complex number and hence express in polar form:

$\sin {120}^{\circ }-i\cos {120}^{\circ }$

Answer 1

Let $z=\sin {120}^{\circ }-i\cos {120}^{\circ }$

$\Rightarrow z=\sin \frac{2\pi }{3}-i\cos \frac{2\pi }{3}$

$\Rightarrow z=\sin (\frac{2\pi }{3}-i\cos \frac{2\pi }{3})$

$\Rightarrow z=\sin (\frac{\pi }{2}+\frac{\pi }{6})-i\cos (\frac{\pi }{2}+\frac{\pi }{6})$

$\Rightarrow z=\cos \frac{\pi }{6}-i(-sin\frac{\pi }{6})$

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

Comparing with $z=|z|\left[\cos arg\right(z)+i\sin arg(z\left)\right]$

$|z|=1\text{and}arg\left(z\right)=\frac{\pi }{6}$

Therefore, Polar form of $z\text{is}\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}$