Solving $z^6=64$ 
$$z^6=64$$

My attempt:
$$\stackrel{\text{Euler}}{=}[r(\cos(\theta)+i\sin(\theta))]^{6}=64$$
$$\stackrel{\text{De moivre}}{=}r^6(\cos(6\theta)+i\sin(6\theta))=64$$
$$\stackrel{\text{Euler}}{=}r^6e^{i6\theta}$$
I know that I should get $6$ roots..
 A: What you did is a good start, you just need to continue. Indeed writing $z=re^{i\theta}$, we find that $r^6e^{i6\theta}=64$, from which we conclude that $r=\sqrt[6]{64}=2$, so our solutions lie on the circle with radius 2 centered at the origin (in the complex plane). Since $64=64e^{0\cdot i}$, comparing the angles we have $6\theta=0+k\cdot2\pi$, where $k\in\{0,1,2,3,4,5\}$. I think you can finish it from here and get your six roots. Note that this works in general: if you have $z^n=\alpha$ for some $\alpha\in\mathbb{C}$, write the right-hand side as $|\alpha|e^{i\arg(\alpha)}$ and the left-hand side as $re^{i\theta}$, then $r=|\alpha|$ and $n\theta=\arg(\alpha)+k\cdot2\pi$, where $k\in\{0,1,\ldots,n-1\}$.
A: $$z^6=64\Longleftrightarrow$$
$$z^6=\left|64\right|e^{\arg\left(64\right)i}\Longleftrightarrow$$
$$z^6=64e^{oi}\Longleftrightarrow$$
$$z=\left(64e^{2\pi ki}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$z=2e^{\frac{1}{6}\cdot 2\pi ki}\Longleftrightarrow$$
$$z=2e^{\frac{1}{3}\cdot \pi ki}$$
With $k\in\mathbb{Z}$ and $k:0-5$

So the solutions are:
$$z_0=2e^{\frac{1}{3}\cdot \pi \cdot 0 i}=2$$
$$z_1=2e^{\frac{1}{3}\cdot \pi \cdot 1 i}=2e^{\frac{\pi}{3}i}$$
$$z_2=2e^{\frac{1}{3}\cdot \pi \cdot 2 i}=2e^{\frac{2\pi}{3}i}$$
$$z_3=2e^{\frac{1}{3}\cdot \pi \cdot 3 i}=-2$$
$$z_4=2e^{\frac{1}{3}\cdot \pi \cdot 4 i}=2e^{-\frac{2\pi}{3}i}$$
$$z_5=2e^{\frac{1}{3}\cdot \pi \cdot 5 i}=2e^{-\frac{\pi}{3}i}$$
