Question

What is the trigonometric form of `9 e^( (3pi)/2 i )`?

Answer 1

`9*(cos((3pi)/2)+isin((3pi)/2))`

Explanation:

Recall Eulers formula:
`e^(iz)=cosz+isinz`

In this case `z=(3pi)/2`

Therefore `9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))`

If you wish to simplify this further, `cos((3pi)/2)=0`, and `sin((3pi)/2)=-1`, therefore:

`9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))=9*(0+i*(-1))=-9i`