Convert the given complex number in polar form: $-1-i$

Let $z = -1-i$.
We represent the complex number in polar form as $z = r(\cos \theta + i \sin \theta)$,where $r = |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.
Now,$r \cos \theta = -1$ and $r \sin \theta = -1$.
$\Rightarrow \cos \theta = -\frac{1}{\sqrt{2}}$ and $\sin \theta = -\frac{1}{\sqrt{2}}$.
Since both $\cos \theta$ and $\sin \theta$ are negative,the angle $\theta$ lies in the $III$ quadrant.
The reference angle $\alpha$ is given by $\tan \alpha = |\frac{-1}{-1}| = 1$,so $\alpha = \frac{\pi}{4}$.
In the $III$ quadrant,$\theta = -(\pi - \alpha) = -(\pi - \frac{\pi}{4}) = -\frac{3\pi}{4}$.
Thus,the polar form is $\sqrt{2}(\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4}))$.