Find all the solutions: $z^5=1-i$ 
Find all the solutions: $z^5=1-i$

my solution:
$$z^5=e^{5\log z}, \quad  1-i=e^{\log(1-i)}=e^{\ln(\sqrt2)-i\frac{\pi}{2}+2k\pi i}, k\in \mathbb{Z}$$
$$ \Rightarrow e^{5\ln(|z|)+i5 \text{Arg} z+10k\pi i}=e^{\ln(\sqrt2)-i\frac{\pi}{2}+2k\pi i}$$
For $k=0$, $$ \Rightarrow  |z|=(\sqrt2)^{\frac{1}{5}}, \quad \text{Arg}=\frac{\pi}{10} $$
and because $z=|z|(\cos(\text{Arg} z)+i\sin(\text{Arg} z))$ we get that  all the solutions are $$z_k=(\sqrt2)^{\frac{1}{5}}\left(\cos(\frac{\pi}{10}+2k\pi)+i\sin(\frac{\pi}{10}+2k\pi)\right)$$ for $k=0,1,2,3,4$.
Is my solution correct?
 A: You have$$1-i=\sqrt2e^{\pi i/4}$$and therefore the numbers $z\in\Bbb C$ such that $z^5=1-i$ are$$\sqrt[10]2e^{\pi i/20+2\pi ik/5},$$with $k\in\{0,1,2,3,4\}$, by de Moivre's formula. There is not need to mention logarithms here.
A: Notice that $(1-i)^5=-4(1-i)$ therefore $z_0=-\sqrt[5]{\frac 14}(1-i)$ verifies ${z_0}^5=1-i$.
The other roots are $z_k=w^kz_0$ where $w$ is are fifth root of unity: $\, w^5=1\iff w=\large{e^{\frac{2ik\pi}5}}$
Note that my $z_0$ correspond to $k=3$ in José's answer.
A: I suppose that the intent was to use De Moivre's Theorem, but since the argument of the given quantity is a multiple of 15° we should be able to render an algebraic solution for the fifth roots using only square roots combined with the real radical derived from the common absolute value.
Raise the argument $1-i$ to the fourth power which (for arguments divisible by 15°) will give a real number times a cube root of unity. The latter factor is just $1$ for an argument divisible by $45°$, thus we get the real number $(1-i)^4=-4$.
So then $(1-i)^5=-4(1-i)$ and from this, we have a solution
$(1-i)^{1/5}=-\sqrt[5]{1/4}(1-i)$
The other solutions are products of this with fifth roots of unity. Using $\cos(72°)=(\sqrt5-1)/4$ and $\sin(72°)=\sqrt{1-\cos^2(72°)}$ we then obtain the complete solution
$(1-i)^{1/5}=-\sqrt[5]{1/4}(1-i)\left(\dfrac{\sqrt5-1}{4}+i\sqrt{\dfrac{5+\sqrt5}{8}}\right)^n,n\in\mathbb{Z}$
