How do you find the two square roots of 2i?

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Answer 1 · 2014-10-26T19:21:42

Let $z = r (cos θ + i sin θ)$ $z=r(cos theta+isin theta)$ be the square-roots of $2 i$ $2i$ .

$z^{2} = {[r (cos θ + i sin θ)]}^{2}$ $z^2=[r(cos theta+isin theta)]^2$

by De Moivre's Theorem,

$= r^{2} (cos 2 θ + i sin 2 θ) = 2 i = 2 (0 + 1 i)$ $=r^2(cos2theta+isin2theta)=2i=2(0+1i)$

$\Rightarrow r^{2} = 2 \Rightarrow r = \sqrt{2}$ $Rightarrow r^2=2 Rightarrow r=sqrt{2}$

$Rightarrow{(cos2theta=0),(sin2theta=1):}}Rightarrow2theta=pi/2+2npi Rightarrow theta=pi/4+npi$ ,

where $n$ is any integer.

So,

$z={sqrt{2}[cos(pi/4)+isin(pi/4)],sqrt{2}[cos({5pi}/4)+isin({5pi}/4)]}$

$={sqrt{2}(1/sqrt{2}+1/{sqrt{2}}i),sqrt{2}(-1/sqrt{2}-1/sqrt{2}i)}$

$={1+i,-1-i}$

I hope that this was helpful.