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#

1 Answer
Nov 29, 2017

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#