How do you multiply $e^(( 19pi )/ 12 ) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2018-01-26T11:10:51

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

Apply Euler's Identity

$e^(itheta)=costheta+isintheta$

$i^2=-1$

Therefore,

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

$e^(i19/12pi)=cos(19/12pi)+isin(19/12pi)$

$cos(19/12pi)=cos(5/4pi+1/3pi)$

$=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)$

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

$=-sqrt2/4+sqrt6/4$

$=(sqrt6-sqrt2)/4$

$sin(19/12pi)=sin(5/4pi+1/3pi)$

$=sin(5/4pi)cos(1/3pi)+cos(5/4pi)sin(1/3pi)$

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

$=-sqrt2/4-sqrt6/4$

$=-(sqrt6+sqrt2)/4$

Finally,

$e^(i19/12pi)*e^(ipi/2)=((sqrt6-sqrt2)/4-i(sqrt6+sqrt2)/4)*(i)$

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

How do you multiply e^(( 19pi )/ 12 ) * e^( pi/2 i ) e^(( 19pi )/ 12 ) * e^( pi/2 i ) in trigonometric form?