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

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Answer 1 · 2016-11-04T06:19:17

$e^(pi/4i)*e^(pi/2i)=-1/sqrt2+1/sqrt2i$

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

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

$e^(pi/4i)*e^(pi/2i)$

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

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

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

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

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

= $cos((3pi)/4)+isin((3pi)/4)$

= $-1/sqrt2+1/sqrt2i$

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