Question

Powers (how `2^(2017/2)=sqrt2*2^1008` works)?

I have `(1-i)^2017=?`
I know that `(cistheta)^n=cis(ntheta)`
`=> ... =>`
`=2^1008-i2^1008`

The problem is HOW am I solving:
`2^(2017/2)=sqrt2*2^1008`

Answer 1

First remember that:

`sqrt(a^3)=sqrt(axxa^2)=>asqrta`
`a^(x/y)=root[y] (a^x)`
`sqrt(a^x)=a^(x/2)`

We know that `2^(2017/2) = sqrt(2^2017)`

By our second and third rule, we know that `sqrt(2^2017)=sqrt(2xx2^2016)=>2^(2016/2)sqrt2`

When simplified, it becomes `2^1008sqrt2`