Use CR Equations to show f'(z^n)=nz^n-1

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 ?

Re: Use CR Equations to show f'(z^n)=nz^n-1

Quote:

Originally Posted by

**olski1** if f is entire then the derivative may be writen as f'(z)=du/dr +idv/dr = dv/dx-idu/dx

It should be $\displaystyle f'(z)=(u_r+iv_r)(\cos \theta -i\sin \theta)$ .