Question

How can you use trigonometric functions to simplify `9 e^( ( pi)/3 i )` into a non-exponential complex number?

Answer 1

`9/2+(9sqrt3)/2i`

Explanation:

We recall that, `e^(itheta)=costheta+isintheta`.

Hence, `9e^(ipi/3)=9(cos(pi/3)+isin(pi/3))`

`=9(1/2+isqrt3/2)`

`=9/2+(9sqrt3)/2i`