Question

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

Answer 1

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

Explanation:

`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

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

Explanation:

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)`