Question

Question #d3532

Answer 1

`0+i`

Explanation:

We can solve this using De Moivre's theorem, what states;

`( costheta + isintheta )^n = cosntheta + i sinntheta`

What we will use without proof, but we can prove by induction

So applying this theorem;

`( cos(pi/2) + isin(pi/2) )^17` = `cos((17pi)/2) + isin((17pi)/2)`

So hence apply the knowledge that;

`sin ((npi)/2)` = 1 for `n = {1,5,9,...4k-3}, k in ZZ`
`cos((npi)/2)` = 0 for `n in {1,3,5...2m-1},m in ZZ`

What we can see when sketching these functions graph{sinx [-10, 10, -5, 5]}

graph{cosx [-10, 10, -5, 5]}

As `n = 17` holds for both statements prior;

So hence `cos((17pi)/2) + isin((17pi)/2)` = `0+i`