Solving a complex equation

Question

$$(z+i)^3+2(z+i)^{-3}+2=0$$

What I have manage to come up with is:

$$w=(z+i)^3$$

$$w+2w^{-1}+2=0$$

$$w+\frac{2}{w}+2=0$$

$$w^2+2w+2=0$$

$$w_{1,2}=\frac{-2\pm\sqrt{4-8}}{2}=\frac{-2\pm2i}{2}=-1\pm i$$

$$(z+i)^3=-1\pm i$$

How to continue?

Answer 1

HINT:

Now $-1-i=\sqrt2e^{i(\pi+\pi/4)}$

and $-1+i=\sqrt2e^{i(\pi-\pi/4)}$

Now use this and How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?

Answer 2

Using polar notation we have: $$ -1+i=\sqrt{2}e^{\frac{3\pi i}{4}} \qquad -1-i=\sqrt{2}e^{\frac{5\pi i}{4}} $$ so, substituting this two values in $(z+i)^3$ and taking all the third-roots we find: $$ z+i=\sqrt[6]{2}e^{(\frac{3}{4}+2k)\frac{\pi}{3}i} \qquad z+i=\sqrt[6]{2}e^{(\frac{5}{4}+2k)\frac{\pi}{3}i} $$

that, for $k\in \{0,1,2\}$ gives $2\times 3$ different values of $z+i$, and from these we find the six values of $z$.