Question 965200
{{{i=0+i=cos(90)+isin(90)}}}
So then,
{{{sqrt(i)=(cos(90)+isin(90))^(1/2)}}}
Using DeMoivre's theorem,
{{{sqrt(i)=cos((90+k*(360))/2)+isin((90+k*(360))/2)}}} for k=0 and k=1.
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{{{k=0}}}
{{{sqrt(i)=cos(90/2)+isin(90/2)}}}
{{{sqrt(i)=cos(45)+isin(45)}}}
{{{sqrt(i)=(sqrt(2))/2)(1+i)}}}
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{{{k=1}}}
{{{sqrt(i)=cos(450/2)+isin(450/2)}}}
{{{sqrt(i)=cos(225)+isin(225)}}}
{{{sqrt(i)=-(sqrt(2)/2)(1+i)}}}