Question

How do you evaluate `e^( ( 5 pi)/4 i) - e^( ( 3 pi)/4 i)` using trigonometric functions?

Answer 1

`sqrt(2)e^((3pi)/2*i)`

Explanation:

`e^(itheta) = cos(theta)+isin(theta)` so:

`e^((5pi)/4i) = cos((5pi)/4)+isin((5pi)/4) = -sqrt(2)/2-sqrt(2)/2i`
and
`e^((3pi)/4i) = cos((3pi)/4)+isin((3pi)/4) = -sqrt(2)/2+sqrt(2)/2i`

Which means that

`e^((5pi)/4i) - e^((3pi)/4i) = -sqrt(2)/2-sqrt(2)/2i - (-sqrt(2)/2+sqrt(2)/2i)`

`=-sqrt(2)/2-sqrt(2)/2i +sqrt(2)/2-sqrt(2)/2i`

`=-sqrt(2)i`

which is a complex number with `r = sqrt(2)` and `theta = (3pi)/2`

so

`e^((5pi)/4i) - e^((3pi)/4i) = sqrt(2)e^((3pi)/2*i)`