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

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Answer 1 · 2017-09-25T14:18:47

The answer is $=1$

We apply Euler's identity

$e^(itheta)=costheta +i sin theta$

$i^2=-1$

$e^a *e^b=e^(a+b)$

So,

$e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+1*i=i$

$e^(i3pi/2)=cos(3/2pi)+isin(3/2pi)=0-1*i=-i$

Therefore,

$e^ (ipi/2)* e^(i3pi/2)=i*-i=-i^2=1$

We can also perform

$e^(ipi/2)*e^(i3pi/2)=e^((ipi/2+i3pi/2))=e^(i4pi/2)=e^(i2pi)$

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

$=1+i*0=1$

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