So the first part to this question was,
Use the CR equations to show that z^n is entire for any n using polar coordinates.
hence,
du/dr=n(r^n-1)(cosnx) dv/dr=n(r^n-1)(sinnx)
du/dx=(-r^n)(nsinnx) , dv/dx=(r^n)(ncosnx).
Which via the CR equalities, are differentiable for all C, for any n => Entire
The second part to the question is, using your answer above show that
(z^n)'=nz^n-1
Okay I have two problems with this. Firstly,
if f is entire then the derivative may be writen as
f'(z)=du/dr +idv/dr = dv/dx-idu/dx
=> f'(r,x)=nr^(n-1)(cosnx+isinnx)= nr^n(cosnx+isinnx)
First off, they are not the same! so why does this not hold true.
Secondly if , f'(r,x)=nr^(n-1)(cosnx+isinnx), is correct.
i cannot see how it equates to (z^n)'=nz^n-1 ?