<s>[INST] Understand roots of $(z+i)^2=(\sqrt3+i)^3$ I'm trying to solve the equation
$$(z+i)^2=(\sqrt3+i)^3$$ but I don't know how to extract
the roots
$$(z+i)^2=(\sqrt3+i)^3 \rightarrow (z+i)^2=8i \rightarrow z^2+(2i)z-(8i+1)=0$$
$z_{1,2}=-i \pm \sqrt{8i}$.
According to my book the solutions are $2+i$ and $-2-3i$
and I think that they are another way to write them but I don't know how to have them.
Can someone help me to understand?
 [/INST] You have $z_{1,2} = -i + K$ where $K^2 = 8i$.
So find $K$ where $K^2 = 8i$.....
Let $K = a+bi$ so $K^2 = (a^2 -b^2) + 2abi = 0 + 8i$ so $a^2-b^2 =0$ and $2ab=8$
So solve for $a$ and $b$.....
$a^2 - b^2 =0$ means $a=\pm b$ and $2ab = 8$ means $ab=4$ means $a$ and $b$ are both the same sign.  So $a=b$ and $ab =a^2=4$ so $a =\pm 2$ and $b=\pm 2$.
So $K = \pm (2 + 2i)$ and $z_{1,2}=-i \pm(2+2i) = \begin{cases}2+i\\-2-3i\end{cases}$.
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