<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] let $$x+iy=\sqrt{8i}$$
$$(x+iy)^2={8i}$$
$$x^2-y^2+2xyi=0+8i$$
compare real and imaginary parts of both sides:
$x^2-y^2=0\;; 2xy=8$
$$x^2-y^2=0$$
square both sides
$$\implies (x^2-y^2)^2 =0\implies (x^2+y^2)^2-(2xy)^2=0$$
$$\implies (x^2+y^2)^2=64 \implies (x^2+y^2)=\pm 8 $$
since sum of square of two real no. can't be negative so
$$x^2+y^2=8\text{ and }x^2-y^2=0$$
solving these two eqn will give $x=y=\pm 2 $
so $$\sqrt {8i}=\pm 2(1+i)\implies 2+2i,-2-2i$$
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