Finding roots using Euler's formula I'm going through past exam papers and have come accross the following question:
find all solutions to $z^5=2-2i $
for this question I was going to use Euler's formula:
$ e^{i(2k+1)\pi}=-1 $
$ -2e^{i(2k+1)\pi}=2 $
$ -2e^{i(2k+1)\pi}-2i=2-2i $
$ z^5 =-2e^{i(2k+1)\pi}-2i$
$ z = -2^{\frac 1 5}e^{\frac {i(2k+1)\pi}{5}}-2i^{\frac 1 5}$
I'll then sub in k=0,1,2,3,4
am I on the right track here?
thanks!
 A: You fell for the freshman's dream:  $(a+b)^n=a^n+b^n$.  
So, instead,  write  $2-2i=2\sqrt2e^{\frac{7\pi}4i}$ (using Euler's formula).
Now $z^5=2\sqrt2e^{\frac{7\pi}4i}$.
So $z=\sqrt[10]{8}e^{\frac{7\pi}{20}i}$ is one solution. 
There are $4$ more.
Take a primitive $5$th root of unity,  $\rho=e^{\frac{2\pi i}5}$.  Then the other $4$ are:  $\rho z,\rho^2z,\rho^3z$ and $\rho^4z$.
That is,  $\sqrt[10]8e^{(\frac{7\pi+8\pi k}{20})i}$, for $k=0,1,2,3$ and $4$.
A: Using the Euler's Formula is a good approach but I would use:
$$z=re^{ix} \rightarrow z^5=r^5e^{5xi}$$
$$2-2i=2\sqrt{2}e^{(7\pi/4+2k\pi)i}$$
Now compare both:
$$r^5e^{5xi}=2\sqrt{2}e^{(7\pi/4+2k\pi)i}$$
Once both complex numbers are equal then they must have the same magnitude ($r^5=2\sqrt{2}$) and also $e^{5xi}=e^{(7\pi/4+2k\pi)i}$, so:
\begin{cases}
r^5=2\sqrt{2}\\
e^{5xi}=e^{(7\pi/4+2k\pi)i}
\end{cases}
P.S: Remember that in general if we have $u=r_1e^{ix_1}$\ $v=r_2e^{ix_2}$ and if $z=w$ then we must have $r_1=r_2$ and $e^{ix_1}=e^{ix_2}$.
Can you finish?
A: We can write $2-2i$ as:
$z^5=2-2i=2\sqrt2e^{i{-\pi\over4}}=2\sqrt2e^{i{-\pi\over4}+2k\pi}$
now we take the root:
$z=(2\sqrt2)^{1\over5}e^{i{-\pi\over20}+{2k\pi\over5}}$
with $k\in \mathbb Z$, $k=0,1,2,3,4$.
Your approach fails because $z=(2-2i)^{1\over5}\ne(2)^{1\over5}-(2i)^{1\over5}$
A: Hint:
$$e^{7\pi/4}=\frac1{\sqrt 2}-i\frac1{\sqrt 2}$$
Therefore 
$$2-2i=2\sqrt 2e^{7\pi/4}$$
