How do you multiply $e^(( 5 pi )/ 4 i) * e^( pi/2 i )$ in trigonometric form?

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Answer 1 · 2016-04-13T12:35:50

$e^((5pi/4)i)*e^((pi/2)i)=-1/sqrt2-i1/sqrt2$

A complex number can be written in polar form in two ways - either as $r*e^(itheta)$ or as $rcostheta+irsintheta$ .

Hence,

$e^((5pi/4)i)=cos(5pi/4)+isin(5pi/4)$ and

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

Hence, $e^((5pi/4)i)*e^((pi/2)i)$

= $(cos(5pi/4)+isin(5pi/4))*(cos(pi/2)+isin(pi/2))$

= $(cos(5pi/4)+isin(5pi/4))*(0+i)$

as $cos(pi/2)=0$ and $sin(pi/2)=1$

= $(icos(5pi/4)+i^2sin(5pi/4))$

= $-sin(5pi/4)+icos(5pi/4)$

= $-1/sqrt2-i1/sqrt2$

How do you multiply e^(( 5 pi )/ 4 i) * e^( pi/2 i ) e^(( 5 pi )/ 4 i) * e^( pi/2 i ) in trigonometric form?