Question

How do you evaluate `e^( ( pi)/2 i) - e^( ( 23 pi)/3 i)` using trigonometric functions?

Answer 1

With the help of Euler's formula we can take this right down to
0.5 + `i` (1 + `(sqrt3)/2`)

Explanation:

Start with Euler's formula: `e^(ix)` = `cos x + i Sin x`

Your expression becomes:

cos `pi/2` + `i` sin `pi/2` - cos `(23pi)/3` - `i` sin `(23pi)/3`

To simplify, note cos `pi/2` = 0, sin `pi/2` = 1

and since each complete revolution of a unit circle is 2`pi`, which we will write as `(6pi)/3`, cos `(23pi)/3` = cos `(5pi)/3`
and likewise for sin `(23pi)/3` = sin `(5pi)/3`

So, we now have the original expression simplified down to

`i` - cos `(5pi)/3` - `i` sin `(5pi)/3`

where cos `(5pi)/3` = 0.5 and sin `(5pi)/3` = - `(sqrt3/2)`

So, finally, the result is

0.5 + `i` (1 + `(sqrt3)/2`)