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?
 A: How do you get your solutions? They are wrong.
The Euler formula is
$$
e^{ti}=\cos(t)+\sin(t)i.
$$
Since $i=e^{\frac12\pi i+2k\pi i}=e^{\left(2k+\frac12\right)\pi i}$ for all $k\in\mathbb{Z}$ we get
$$
z^3=i=e^{\left(2k+\frac12\right)\pi i} \Leftrightarrow z=e^{\left(\frac23k+\frac16\right)\pi i}.
$$

Now you see that you get different solutions for $k\in\{0,1,2\}$ which are $$z_0=e^{\frac16\pi i},~z_1=e^{\frac56\pi i}\text{ and }z_2=e^{\frac32\pi i}.$$ Using the Euler formula we get \begin{align}z_0&=\cos\left(\frac16\pi\right)+\sin\left(\frac16\pi\right)i=\frac{\sqrt{3}}2+\frac12i\\z_1&=\cos\left(\frac56\pi\right)+\sin\left(\frac56\pi\right)i=-\frac{\sqrt{3}}2+\frac{1}2i\\z_2&=\cos\left(\frac32\pi\right)+\sin\left(\frac32\pi\right)i=-i\\\end{align}

A: 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:
$$z=\cos\frac{\frac{\pi}{2}+2\pi k}{3}+i\sin\frac{\frac{\pi}{2}+2\pi k}{3}=e^{i\frac{(\frac{\pi}{2}+2k\pi)}{3}}=e^{i(\frac{\pi}{6}+\frac{2k\pi}{3})}$$
A: My hint: write $z$ in polar form, i.e. $z=re^{i\theta}$; from this we have $z^3=r^3e^{i3\theta}$.
Then write $i=e^{i\frac{\pi}2}=e^{i\frac{\pi}2+2k\pi i}$ from the periodicity of the complex exp.
Thus
\begin{align*}
z^3=i
&\Longleftrightarrow (r=1)\;\; \wedge\;\;(3\theta=\pi/2+2k\pi)\\
&\Longleftrightarrow z=e^{i(\frac{\pi}{6}+\frac23k\pi)}.
\end{align*}
Finally let $k\in\Bbb Z$ and check what are the different results you get!
