Solving $z^3$ Using Euler Formula I know that $z^3=i$ can be solve by:
$$z=cos0+isin\frac{\frac{\pi}{2}+2\pi k}{3}$$ where $k\in\{0,1,2\}$
So we get:
$z_0=isin\frac{\pi}{6}$
$z_1=isin\frac{5\pi}{6}$
$z_2=isin\frac{9\pi}{6}$
Using Euler formula we just write it as $$z^3=e^{\frac{\pi}{2}i}$$ and then?

The exponential function in $\mathbb{C}$ is periodic with period $2\pi i$, so the equation $z^3=i$ in exponential form becomes:
$$
z^3=e^{i(\frac{\pi}{2}+2k\pi)}
$$
and we have:
$$
z=\left(z^3\right)^\frac{1}{3}=e^{i(\frac{\pi}{6}+\frac{2k\pi}{3})}
$$
and this gives three different solutions:
$$
k=0 \quad \rightarrow \quad z=e^{i\frac{\pi}{6}}
$$
$$
k=1 \quad \rightarrow \quad z=e^{i(\frac{\pi}{6}+\frac{2}{3}\pi)}=e^{i\frac{5\pi}{6}}
$$
$$
k=2 \quad \rightarrow \quad z=e^{i(\frac{\pi}{6}+\frac{4}{3}\pi)}=e^{i\frac{9\pi}{6}}
$$
Note that your use of the Euler formula is wrong. The correct result is:
$$z=\cos\frac{\frac{\pi}{2}+2\pi k}{3}+i\sin\frac{\frac{\pi}{2}+2\pi k}{3}=e^{i\frac{(\frac{\pi}{2}+2k\pi)}{3}}=e^{i(\frac{\pi}{6}+\frac{2k\pi}{3})}$$
