How do you multiply $e^(( 9 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-09-26T10:23:14

As $e^(itheta)=costheta+isintheta$ , we have

$e^((9pi)/4i)=cos((9pi)/4)+isin((9pi)/4)$ and

$e^(pi/2i)=cos(pi/2)+isin(pi/2)$

Hence $e^((9pi)/4i)*e^(pi/2i)=(cos((9pi)/4)+isin((9pi)/4))(cos(pi/2)+isin(pi/2))$

= $cos((9pi)/4)(cos(pi/2)+isin(pi/2))+isin((9pi)/4))(cos(pi/2)+isin(pi/2))$

= $cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)+i^2sin((9pi)/4)sin(pi/2))$

= $cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))$

= $(cos((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))+i(sin((9pi)/4)cos(pi/2)+cos((9pi)/4)sin(pi/2))$

= $cos(((9pi)/4)+(pi/2))+isin(((9pi)/4)+(pi/2))$

= $cos((11pi)/4)+isin((11pi)/4)$

= $e^((11pi)/4)$

How do you multiply e^(( 9 pi )/ 4 i) * e^( pi/2 i ) e^(( 9 pi )/ 4 i) * e^( pi/2 i ) in trigonometric form?