Express $(1-i)^{11}$ in cartesian form.

Hint: $1-i=\sqrt 2e^{-i\pi/4}{}{}{}{}$.

To find the above equality you can think about it geometrically and use the known techniques:

Or algebraically by proving that given any $a,b\in \mathbb R$, it holds that $a+ib=\sqrt{a^2+b^2}e^{i\arg(a+ib)}$, where $$\arg(a+ib)=\begin{cases} \arctan(b/a), &\text{if }a>0\\ \pi /2, &\text{if }a=0\land b>0\\ \arctan(b/a)+\pi, &\text{if }a<0\\ -\pi/2, &\text{if }a=0\land b<0\end{cases}$$

Hint: $(1-i)^2=-2i{}{}{}{}{}{}{}{}$.

Expanding expressions like this one is not easy. Therefore, it is useful to rewrite it to its polar form. We know that the angle is $\varphi=-\frac 14 \pi$ and the length is $r=\sqrt 2$. We can find $r$ by Pythagoras' theorem: $$ r=\sqrt{1^2+(-1)^2}=\sqrt 2 $$ and we can find $\varphi$ by just looking at the point. (We know that $x=-y$, so the angle is of the form $k\frac 12\pi+\frac 14\pi$ for some integer $k$, and because it goes to the lower right, it has to be $-\frac 14\pi$.)

When taking the eleventh power, we get $$ \varphi'=11\varphi=-\frac{11}4\pi=-\frac 34\pi=\frac 54 \pi\\ r'=r^{11}\left(\sqrt2\right)^{11}=2^{\frac 12\cdot 11}=2^{5+\frac 12}=32\sqrt 2 $$ Now, we need to transform the result back to Carthesian form: $$ x=r'\cos\varphi'=-32\\ y=r'\sin\varphi'=-32 $$ Thus, we get the result $-32-32 i$ or $(-32,-32)$.