Question

-64=(2cos(pi/2)+2i sin(pi/2))^n..... n=??​

Answer 1

The real solution is
`n=6`.

Explanation:

`\cos((\pi)/(2))=0` while
`\sin((\pi)/(2))=1`

So the equation becomes:

`-64=(2(0)+2i(1))^n`

`-64=(0+2i)^n`

`-64=(2i)^n`

We know that
`2^6=64`. So let's see what
`n=6` gives us:

`(2i)^6=64i^6=64i^4i^2=64(1)(-1)=-64`.

`-64` is the result we wanted.

`n=6` is therefore a solution.