complex numbers equation with 3rd power I have problem with this equation. Can someone help mi with it?
\begin{align*} z^3+3z^2+3z &amp;= i - 1 \end{align*}
That's what I have:
\begin{align*} (z+1)^3 = i \end{align*}
If ${(z+1)^{3}=i}$, what we essentially want to do is find the cube roots of ${i}$. This is much easier than it sounds. Firstly, we represent ${i}$ in polar form. This is simply
$${e^{i\left(\frac{\pi}{2} + 2n\pi\right)}}$$
This can be seen with an argand diagram (it's just an application of Euler's formula):
$${=cos\left(\frac{\pi}{2}+2n\pi\right)+i\sin\left(\frac{\pi}{2}+2n\pi\right)=i}$$
Now raising this to the one third power gives
$${\Rightarrow z+1=e^{\frac{i}{3}\left(\frac{\pi}{2}+2n\pi\right)}}$$
Simply plugging in ${n=\{0,1,2\}}$ will give the three roots we need,
${z+1=e^{\frac{i\pi}{6}}, e^{\frac{5i\pi}{6}},e^{\frac{3\pi i}{2}}}$
(plugging these again into Euler's formula and subtracting $1$ will give you your answers for $z$).
Yes, that is correct. And, since $i=e^{\pi i/2}$, $z+1$ is one of these three numbers:

$e^{\pi i/6}\left(=\frac{\sqrt3}2+\frac i2\right)$,
$e^{5\pi i/6}\left(=-\frac{\sqrt3}2+\frac i2\right)$,
$e^{9\pi i/6}(=e^{3\pi i/2}=-i)$.