Question

The polar form of complex number $-3\sqrt{2}+3\sqrt{2}i$ is

• A. $6(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$
• B. $3(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})$
• C. $6(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4})$
• D. $3(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Answer 1

The correct option is A $6(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Let $z=-3\sqrt{2}+3\sqrt{2}i$
Then,
$r=|z|=\sqrt{(-3\sqrt{2}{)}^2+(3\sqrt{2}{)}^2}=6\tan \alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=1\Rightarrow \alpha =\frac{\pi }{4}$
Since the point representing $z$ lies in the second quadrant, therefore, the argument of $z$ is given by
$\theta =\pi -\alpha =\pi -\left(\frac{\pi }{4}\right)=\left(\frac{3\pi }{4}\right)$
So, the polar form of $z=-3\sqrt{2}+3\sqrt{2}i$ is
$z=r(\cos \theta +i\sin \theta )\Rightarrow z=6(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$