How do you multiply $e^(( 3 pi )/ 2 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-03-27T08:35:51

$e^((3pi)/2i)*e^((3pi)/2i)=-1$

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

$e^((3pi)/2i)=cos((3pi)/2)+isin((3pi)/2)$ .

Hence, $e^((3pi)/2i)*e^((3pi)/2i)=$

${cos((3pi)/2)+isin((3pi)/2)}*{cos((3pi)/2)+isin((3pi)/2)}$ or

= ${cos^2((3pi)/2)+i^2sin^2((3pi)/2)+2isin((3pi)/2)cos((3pi)/2)}$

= ${cos^2((3pi)/2)+(-1)sin^2((3pi)/2)+i(2sin((3pi)/2)cos((3pi)/2))}$

= ${(cos^2((3pi)/2)-sin^2((3pi)/2))+i(2sin((3pi)/2)cos((3pi)/2))}$

= $cos(2xx(3pi)/2)+isin(2xx(3pi)/2)$

= $cos3pi+isin3pi$

= $-1+ixx0=-1$

How do you multiply e^(( 3 pi )/ 2 i) * e^( 3 pi/2 i ) e^(( 3 pi )/ 2 i) * e^( 3 pi/2 i ) in trigonometric form?