I want to solve: $(z+2-i)^6=27i$
My thought was expressing it as:
$(z+2-i)^6=27 e^{i(\frac{\pi}{2}+ 2\pi k)}$ where $k \in \mathbb{Z}$
$(z+2-i)=\pm \sqrt{3} e^{i(\frac{\pi}{12}+ \frac{1}{3}\pi k)}$ But if I now try to write this in standard form It will look horrible, is there something in my approach that I'm tackling incorrectly or inefficiently?
I get: $z= \pm\sqrt{3} \cos(\frac{\pi}{12}+ \frac{1}{3}\pi k)-2 +i(\pm\sqrt{3} \sin(\frac{\pi}{12}+ \frac{1}{3}\pi k) +1) $
No need to use the $\pm$ sign as $k=3+j$ provides the negative of the solution $k=j$ aswell. The way you dealt with the problem is fine. The space of solutions is well defined.
An alternative solution:
Write $w:=(z+2-i)^2$ and solve
$$w^3=27i$$ or $$w=-3i\omega^k$$ where $\omega$ is a complex cubic root of unity.
Then
$$z+2-i=\pm\sqrt 3\frac{1-i}{\sqrt2}\omega'^k$$ where $\omega'$ is a sixth root of unity, $\text{cis }\dfrac{k\pi}6$.
You get a closed-form expression.