# Complex analysis proof

1. Jun 15, 2009

### daudaudaudau

1. The problem statement, all variables and given/known data
Given |z|<1 and n a positive integer prove that
$$\left|\frac{1-z^n}{1-z}\right|\le n$$

3. The attempt at a solution
I try to find the maximum of the function by differentiation

$$\frac{d}{dz}\frac{1-z^n}{1-z}=\frac{-nz^{n-1}*(1-z)+(1-z^n)}{(1-z)^2}=0\Rightarrow (1-z^n)=nz^{n-1}*(1-z)$$

I then plug this in

$$\left|\frac{nz^{n-1}*(1-z)}{1-z}\right|=n\left|z^{n-1}\right|\le n$$

I guess this works. Does someone have another way to prove it?

2. Jun 15, 2009

$$\left|\frac{1-z^n}{1-z}\right|=\left|\sum_{k=0}^{n-1}z^k\right|\le \sum_{k=0}^{n-1}|z|^k\le n$$