How do you multiply $e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )$ in trigonometric form?

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Answer 1 · 2017-10-15T19:42:32

$= e^(((31pi)/8)i)$

$e^(((3pi)/8)i) * e^ (3(pi/2)i$
$e^(((3pi)/8)i) * e^(((7pi)/2)i)$

$e^(((3pi)/8)i)* e^ (((28pi)/8)i)$
$e^ (((3pi + 28pi)/8)(i+i) = e^(((31pi)/8)i)$

Answer 2 · 2017-10-15T20:53:31

$cos(pi/8)-isin(pi/8)$ or $-(-1)^(7/8)$

Are you sure its $e^(3pi/2)$ not $e^((3pi)/2)$ because it makes more sense when you write it in trig form.

Remember this beauty?
$e^(ipi)=-1$
That was Euler's identity. This is the generalized formula:
$e^(ix)=cosx+isinx$

Therefore, we can break down the two terms involved:

$e^(i3/8pi)=cos((3pi)/8)+isin((3pi)/8)$

$e^(i7/2pi)=cos((7pi)/2)+isin((7pi)/2)=0+i*(-1)=-i$

$e^(i3/8pi)=(e^(i3/2pi))^(1/4)=(-i)^(1/4)$
$e^(i3/8pi)*e^(i3/2pi)=(e^(i3/2pi))^(1/4)*e^(i7/2pi)=(e^(i3/2pi))^(5/4)=(-i)^(5/4)=-(-1)^(7/8)$

Or

$e^(i3/8pi)*e^(i7/2pi) = sin((3pi)/8)-icos((3pi)/8)$

$=cos(pi/8)-isin(pi/8)$

How do you multiply e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i ) e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i ) in trigonometric form?