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

1 Answer
Nov 27, 2017

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