Solve for $z$ (complex numbers) $$
z^3=i
$$
The problem simply states to solve for $z$, but I know that there is some concept to be practiced here about the nth roots of unity. I'm just beginning to learn this concept so I didn't really know how to go about solving this, but below is my attempt:
$Let \ z=x+iy$
$$
z=i^{\frac{1}{3}} = -i
$$
$$
r=\sqrt{0^2+(-1)^2}=1
$$
$$
\theta=-\frac{\pi}{2}
$$
$$
z=(1)e^{i(\frac{-\pi}{2})}=e^{\frac{-\pi}{2}i}
$$
 A: This answer was meant for the original problem, the one that had $z^3=1$, the new one can be solved in the exact same manner, replacing $1\mapsto i$.

This method is ugly but requires the fewest theoretical knowledge.
Let $z=x+iy. x,y\in\Bbb R$, then
$$(x+iy)^3=1\iff (x^3-3xy^2)+(3x^2y-y^3)i=1$$
Then
$$
x^3-3xy^2=1\\
3x^2y-y^3=0\iff 3x^2=y^2
$$
Replace in first
$$
x^3-3x(3x^2)=1\iff -8x^3=1\iff x=-\frac 1 2
$$
Then $$y^2=\frac 3 4\rightarrow y=\pm\frac {\sqrt{3}} 2$$
A: Hint:
Use De Moivre's formula.
$z^3 = 1cis(\pi /2)$
$z_k = cis(\pi / 6 +2\pi k / 3)$ Solve for $k=0,1,2$
A: They are said to be cube roots of unity so $z^3-1=0$ so $(z-1)(z^2+z+1)=0$ solving $z=1,\omega ,\omega ^2$ where $\omega =\frac{-1+{\sqrt{3}i}}{2}$ and $\omega^2 = \frac{-1-{\sqrt{3}i}}{2}$
A: $$z^3=i\Longleftrightarrow$$
$$z^3=|i|e^{\arg(i)i}\Longleftrightarrow$$
$$z^3=e^{\frac{\pi i}{2}}\Longleftrightarrow$$
$$z=\left(e^{\left(2\pi k+\frac{\pi}{2}\right)i}\right)^{\frac{1}{3}}\Longleftrightarrow$$
$$z=e^{\frac{1}{3}\left(2\pi k+\frac{\pi}{2}\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-2$

So the solutions are:
$$z_0=e^{\frac{1}{3}\left(2\pi\cdot0+\frac{\pi}{2}\right)i}=e^{\frac{\pi i}{6}}$$
$$z_1=e^{\frac{1}{3}\left(2\pi\cdot1+\frac{\pi}{2}\right)i}=e^{\frac{5\pi i}{6}}$$
$$z_2=e^{\frac{1}{3}\left(2\pi\cdot2+\frac{\pi}{2}\right)i}=-i$$
