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 theta+isin theta)# be the square-roots of #2i#.

#z^2=[r(cos theta+isin theta)]^2#

by De Moivre's Theorem,

#=r^2(cos2theta+isin2theta)=2i=2(0+1i)#

#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.