How can you use trigonometric functions to simplify $5 e^( ( 13 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2017-08-11T09:09:51

The answer is $=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)$

We apply Euler's relation

$e^(itheta)=costheta+isintheta$

$5e^(13/12ipi)=5(cos(13/12pi)+isin(13/12pi))$

$cos(13/12pi)=cos(1/3pi+3/4pi)$

$=cos(1/3pi)cos(3/4pi)-sin(1/3pi)sin(3/4pi)$

$=-1/2*sqrt2/2-sqrt3/2*sqrt2/2$

$=-(sqrt6+sqrt2)/4$

$sin(13/12pi)=sin(1/3pi+3/4pi)$

$=sin(1/3pi)cos(3/4pi)+cos(1/3pi)sin(3/4pi)$

$=-sqrt3/2*sqrt2/2+1/2*sqrt2/2$

$=(sqrt2-sqrt6)/4$

Therefore,

$5e^(13/12ipi)=-5/4(sqrt6+sqrt2)+i5/4(sqrt2-sqrt6)$

How can you use trigonometric functions to simplify 5 e^( ( 13 pi)/12 i ) 5 e^( ( 13 pi)/12 i ) into a non-exponential complex number?