Question

What is the trigonometric form of `5e^(pi i)`?

Answer 1

`5 [cos(pi) + isin(pi) ]`

Explanation:

Euler's theorem states that : `e^(itheta) = costheta + isintheta`

`rArr5e^(pi i) = 5[ cos(pi) + isin(pi) ]`

Note : since `cos(pi) = - 1 and sin(pi) = 0`

the expression may be simplified to - 5