How do you multiply $e^(( 7 pi )/ 12 ) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2017-11-22T14:43:55

The answer is $=-(sqrt2+sqrt6)/4+(sqrt2-sqrt6)/4*i$

Apply Euler's relation

$e^(itheta)=costheta+isintheta$

Therefore,

$e^(i7/12pi)=cos(7/12pi)+isin(7/12pi)$

$cos(7/12pi)=cos(1/4pi+1/3pi)=cos(1/4pi)cos(1/3pi)-sin(1/4pi)sin(1/3pi)$

$=(sqrt2/2).(1/2)-(sqrt2/2)(sqrt3/2)$

$=(sqrt2-sqrt6)/4$

$sin(7/12pi)=sin(1/4pi+1/3pi)=sin(1/4pi)cos(1/3pi)+cos(1/4pi)sin(1/3pi)$

$=(sqrt2/2).(1/2)+(sqrt2/2)(sqrt3/2)$

$=(sqrt2+sqrt6)/4$

$e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+i.1=i$

$i^2=-1$

Therefore,

$e^(i7/12pi).e^(ipi/2)=(((sqrt2-sqrt6)+(sqrt2+sqrt6)i)/4)(i)$

$=-(sqrt2+sqrt6)/4+(sqrt2-sqrt6)/4*i$

How do you multiply e^(( 7 pi )/ 12 ) * e^( pi/2 i ) e^(( 7 pi )/ 12 ) * e^( pi/2 i ) in trigonometric form?