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

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Answer 1 · 2018-03-02T07:04:09

The answer is $=sqrt2/2(1-i)$

Apply Euler's Identity

$e^(itheta)=costheta+isintheta$

$i^2=-1$

Therefore,

$e^(pi/4i)=cos(pi/4)+isin(pi/4)=sqrt2/2+isqrt2/2=sqrt2/2(1+i)$

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

So,

$z=e^(pi/4i)*e^(3/2pii)=sqrt2/2(1+i)*(-i)=sqrt2/2(-i-i^2)$

$=sqrt2/2(1-i)$

$"Verification"$

$z=(cosphi+isinphi)=sqrt2/2(1-i)$

$cosphi=sqrt2/2$

$sinphi=-sqrt2/2$

$phi=-pi/4$ , $[2pi]$

$z=e^(pi/4i)*e^(3/2pii)=e^((pi/4+3/2pi)i)=e^(7/4pii)=e^(-1/4pii)$

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