Answer
$3\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
Work Step by Step
We are given:
$z=-3+3i$
To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$:
$r=\sqrt{(-3)^{2}+3^{2}}=3\sqrt{2}$
To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$:
$\displaystyle \tan\theta=\frac{3}{-3}=-1$
And since $z$ lies in the second quadrant, we have:
$\displaystyle \theta=\frac{3\pi}{4}$
To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$:
$3\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$