Is $(1-i)^5 = 2-2i$? The question asks: 
Is this true that: 
$$(1-i)^5 \stackrel{?}=  2-2i$$

I want to know if I did it right:
$$z = (1-i)$$
$$r = \sqrt{1^2+(-1)^2} = \sqrt{2}$$
$$\tan \theta = \frac{-1}{1} = -1 \Rightarrow \theta = -45$$
$$\Downarrow$$
$$z = \sqrt{2}(\cos -45 + i \sin -45)$$
$$z^5 = (\sqrt{2})^5(\cos (5*-45) + i \sin (5*-45))$$
$$\Downarrow$$
$$(1-i)^5 = (\sqrt{2})^5(\cos (-225) + i \sin (-225)) = -4+4i$$
Therefore: 
$$(1-i)^5 \neq 2-2i$$
Am I right? 
 A: Your solution is correct. In your particular problem, it might have been faster to simply compare the norms of both numbers, but to be fair, if these had been equal, you would've gotten no information out of it.
Here's another approach. Suppose otherwise. Then, dividing both sides of the equation by $1-i$, we get $$(1-i)^4=2.$$ However, by roots of unity, it is easy to verify that the only solutions to $x^4=2$ are $$\sqrt[4]{2},-\sqrt[4]{2},\sqrt[4]{2}i,-\sqrt[4]{2}i.$$ $1-i$ is clearly none of these numbers, contradiction.
A: If you use the binomial theorem you have:
$$(1-i)^5 =\sum _{k=0}^5\binom{5}{k}\cdot 1^{\left(5-k\right)}\left(-i\right)^k=-4+4i\neq 2-2i$$
A: From the polar form of a complex number we know that $(re^{2\pi i\theta})^n=r^ne^{2\pi i (n \theta)}$. Since the radius of $1-i$ is $\sqrt{2}$ we see that $r^5=(\sqrt{2})^5=4\sqrt{2}$. Since the radius of $2-2i$ is $2\sqrt{2}$ it cannot be the correct answer.
A second way to see this is to first note that $1-i$ and $2-2i$ make the same angle with the real axis and this is an eighth of the circle in the clockwise direction. Since taking powers adds the angle to itself so $(1-i)^5$ will only be $\frac{5}{8}$ of the circle in the clockwise direction and so is certainly not in the same direction as $2-2i$ and therefore cannot be the same number.
