Solve $z^5 +32 =0$ 
Solve $z^5 +32 =0$

My attempt : $$z^5 = -32$$
            Multiply the powers on both sides by $\frac{1}{5}$
we get $$z = 2 * (-1)^\frac{1}{5}$$
Now I'm stuck at this step I don't know how to proceed. Kindly help.
 A: Use polar coordinates:
$$z^5=32\cdot(-1)=2^5e^{\pi i+2k\pi i}=2^5e^{\pi i(1+2k)}\;,\;\;k\in\Bbb Z\implies$$
$$\implies z_k:=2\,e^{\frac{\pi i}5(1+2k)}\;,\;\;k=0,1,2,3,4$$
Check the above, and fill in details: why in the last line it is enough to take $\;k=0,1,2,3,4\;$ ? You may to check what you get with $\;k=5,6\;$ , say. Are all the $\;z_k$'s defined above different?
A: We have $$\left(-\dfrac z2\right)^5=1$$
Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
If $x^5-1=0,$
either $x-1=0\iff x=1$
or $x^4+x^3+x^2+x+1=0$
Like  Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
as $x\ne0,$ divide both sides by $x^2$  to find $$0=x^2+\dfrac1{x^2}+x+\dfrac1x+1=\left(x+\dfrac1x\right)^2+\left(x+\dfrac1x\right)-1$$
Solve for $x+\dfrac1x$  then solve for $x$
A: $z$ to the power of a fraction (let's define it as $1$ over $y$) is equal to the $y$th root of $z$.
Thus $-1$ to the power of $1/5$ is equal to the fifth root of $-1$, which is $-1$. Solving the equation now from this point is easy:
$$z = 2 \times (-1)^5 = 2 \times -1 = -2$$
Thus
$$z = -2$$
Check the answer by plugging it in:
$$z^5 + 32 = 0$$
$$-2^5 + 32 = 0$$
$$-32 + 32 = 0$$
$$0 = 0$$
A: Since $(-1)^5=-1$,
$(-1)^{\frac15}$ is equal to $-1$ in the real numbers realm.
But consider $(\cos \frac{2k\pi}5 + i\sin \frac{2k\pi}5)^5=1$ so by multiplying the first answer you get all the answer as
$$-2(\cos \frac{2k\pi}5 + i\sin \frac{2k\pi}5),\quad k=0, 1, 2, 3, 4$$
