Find all solutions of $z^5+a^5=0$ The task is as follows:
Find all solutions of $z^5+a^5=0$, where $a$ is a positive real number.

My initial attempt
(which leads nowhere)
My guess is that i'll have to find the 5 5th roots of $-z^5$:
$w_1 = |-z^5|^5(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})) \\
w_1  = |z|(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})$
Then, the other roots are:
$w_2  = |z|(cos(\frac{\theta+2\pi}{5})+isin(\frac{\theta+2\pi}{5}) \\
w_3  = |z|(cos(\frac{\theta+4\pi}{5})+isin(\frac{\theta+4\pi}{5}) \\
w_4  = |z|(cos(\frac{\theta+6\pi}{5})+isin(\frac{\theta+6\pi}{5}) \\
w_5  = |z|(cos(\frac{\theta+8\pi}{5})+isin(\frac{\theta+8\pi}{5})$
but i'm not sure what to do with this new information. Also, I guess $z^5$ have to be a negative real number, since added to $a^5$ its $0$. Again, not sure what to do with this either. 

To make our life a bit easier, start by writing $z=r(\cos\theta+i\sin\theta)$ for some $r>0$ and some real $\theta,$ so that $$-a^5=z^5\\-a^5=r^5(\cos\theta+i\sin\theta)^5\\-a^5=r^5(\cos5\theta+i\sin5\theta)$$ Now, taking the modulus of both sides gives us $$a^5=r^5,$$ so we'll need $r=a,$ and so $z=a(\cos\theta+i\sin\theta).$ Moreover, since $$-a^5=r^5(\cos5\theta+i\sin5\theta)\\-a^5=a^5(\cos5\theta+i\sin5\theta),$$ then $$-1=\cos5\theta+i\sin5\theta,$$ meaning that we need $\theta$ such that $$-1=\cos5\theta\\0=\sin5\theta.$$ This is true if and only if $5\theta$ is an odd integer multiple of $\pi.$ Can you take it from there to find appropriate values of $\theta$ to give $5$ distinct solutions $z$?
