<s>[INST] what is$ \sqrt{8i}$ Very simple question with an answer that I cannot understand:
I have $\sqrt{8i}$, which, I suppose, is the same as $\sqrt{\sqrt{-64}}$.
How come that $2+2i$ is the same as $\sqrt{8i}$?
My understanding is that $\sqrt{8i}$ is the same as:
(a) $\sqrt{2^3i}$
OR
(b) $2\sqrt{2i}$
I'm pretty sure (a) is correct and (b) might also be correct, but how can you get from there to $2+2i$?
Thanks in advance
 [/INST] $(a+bi)^2=(a+bi)(a+bi)=a^2-b^2+2abi\implies \sqrt{a^2-b^2+2abi}=a+bi$
and for conjugate $(a-bi)^2=(a-bi)(a-bi)=a^2+b^2-2abi\implies \sqrt{a^2+b^2-2abi}=a-bi$
Now if we find $a$ and $b$ satisfy appropriate equotation:
For $(a+bi)^2$:
$a^2-b^2+2abi=8i$. This is valid for any $a=b$ where $2ab=8$, or 
$2a^2=8 \implies a=\pm\sqrt{4}=\pm{2} = b$. Here exists two numbers:
First $$\sqrt{8i}=2+2i$$ and second $$\sqrt{8i}=-2-2i$$
For $(a-bi)^2$:
$a^2+b^2-2abi=8i$. Because $a^2 \ge 0$ and $b^2 \ge 0$, there does not exists solution.
</s>