Express $(1-i)^{11}$ in cartesian form. Express $(1-i)^{11}$ in cartesian form.
Apart from expanding the expression, I don't know how to do this. I've looked at the solution and still don't understand how/why it has been done.
 A: 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 $$\arg(a+ib)=\begin{cases} \arctan(b/a), &\text{if }a>0\\ \pi /2, &\text{if }a=0\land b>0\\ \arctan(b/a)+\pi, &\text{if }a<0\\ -\pi/2, &\text{if }a=0\land b<0\end{cases}$$
A: Hint: $(1-i)^2=-2i{}{}{}{}{}{}{}{}$.
A: 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 $(-32,-32)$.
A: Use Euler's Formula:
$(1-i)^{11} = \sqrt {2}^{11}*(\cos7\pi/4 + i\sin7\pi/4)^{11}$
= $\sqrt {2}^{11}*(\cos((7*11\pi)/4) + i\sin(7*11\pi)/4)$
= $\sqrt {2}^{11}*(-\frac{1}{\sqrt {2}}- i\frac{1}{\sqrt {2}})$
= $-32*(1+i)$
Thanks
Satish
