How do you multiply $e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-03-02T10:14:33

$e^(7pi/4i)=(cos(7pi/4)+isin(7pi/4))=1/sqrt2-i(1/sqrt2)$

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

Trigonometric form of $e^(7pi/4i)$ can be written as

$(cos(7pi/4)+isin(7pi/4))$

As $cos(7pi/4)=cos(-pi/4)=cos(pi/4)=1/sqrt2$ and $sin(7pi/4)=sin(-pi/4)=sin(pi/4)=-1/sqrt2$

$e^(7pi/4i)$ can be written as $1/sqrt2-i(1/sqrt2)$

and that of $e^(3pi/2i)$ can be written as

$(cos(3pi/2)+isin(3pi/2))$ and as $cos(3pi/2)=0$ and $sin(pi/2)=-1$ ,

$e^(3pi/2i)$ can be written as $-i$

How do you multiply e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i ) e^(( 7 pi )/ 4 i) * e^( 3 pi/2 i ) in trigonometric form?