Question

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

Answer 1

`e^{i{3pi}/4} * e^{i{3pi}/2} = 1/sqrt2+ i/sqrt2`

Explanation:

From the identity

`e^{itheta} -= cos(theta) + i sin(theta)`

We write

`e^{i{3pi}/4} * e^{i{3pi}/2} = (cos({3pi}/4) + i sin({3pi}/4)) * (cos({3pi}/2) + i sin({3pi}/2))`

`= (-1/sqrt2 + i (1/sqrt2)) * (0 + i (-1))`

`= 1/sqrt2*(-1 + i) * (- i)`

`= 1/sqrt2*(i + 1)`

`= 1/sqrt2+ i/sqrt2`