Question 964603
{{{0+1i=cos(90)+i*sin(90)}}}
So, then using DeMoivre's theorem,
{{{i^(1/5)=cos((90+360(0))/5)+i*sin((90+360(0))/5))}}}
{{{i^(1/5)=cos(18)+i*sin(18)}}}
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{{{i^(1/5)=cos((90+360(1))/5)+i*sin((90+360(1))/5))}}}
{{{i^(1/5)=cos(90)+i*sin(90)}}}
{{{i^(1/5)=i}}}
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{{{i^(1/5)=cos((90+360(2))/5)+i*sin((90+360(2))/5))}}}
{{{i^(1/5)=cos(162)+i*sin(162)}}}
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{{{i^(1/5)=cos((90+360(3))/5)+i*sin((90+360(3))/5))}}}
{{{i^(1/5)=cos(234)+i*sin(234)}}}
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{{{i^(1/5)=cos((90+360(4))/5)+i*sin((90+360(4))/5))}}}
{{{i^(1/5)=cos(306)+i*sin(306)}}}
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