what is the sup and inf of absolute value of $1+z+z^2+ ...+z^n$? What is the supremum and the infimum of the absolute value of $1+z+z^2+ \dots+z^n$ when $z$ is a complex number and $z$ is inside the unit circle on the complex plane, which means $zz^*<1$?
I tried such as when $z \geq 1$ on $\mathbb{R}$ it seems to have the supremum $n+1$ and when $z \geq 0$ it seems to have the infimum $1$,
but I couldn't prove it.
 A: A very elegant approach is to look at the factorization of the polynomial $1+z+z^2+\ldots+z^n$. This obviously has as roots complex $(n+1)^{th}$ roots of unity. Interestingly these lie on the boundary of your domain. So if I write
$$|1+z+z^2+\ldots+z^n| = |z-\zeta_1||z-\zeta_2|\ldots|z-\zeta_{n-1}|.$$
Obviously as you approach one of the roots on the boundary the function will approach zero which is the infimum. What about the supremum? It will be attained somewhere on the boundary $|z|=1$. (Check this post.) 
A: $$1+z+z^2+...+z^n=\frac{z^{n+1}-1}{z-1}$$
It gets pretty easier with this form.
A: $|1+z+z^2 + \dotso + z^n| \leq |1| + |z| + \dotso +|z^n|  = |1| + |z| + \dotso +|z|^n  = n+1 $, ( because |a+b| <= \leq |a|+|b| and |ab|=|a||b| )
if for $z=1$ function has value: $n+1$, that's supremum.
Infimum is $0$ obviously (for $z=0$).
A: $f(z,n)=\sum_{i=1}^nz^i$
Supremum: $|\sum_{i=0}^nz^i|<= \sum_{i=0}^n |z^i| = n+1$.
From the other hand z=1 gives you this value => sup |f(z,n)|=n+1.
Infimum: $\sum_{i=0}^nz^i = \frac{z^{n+1}-1}{z-1}$. Using $z = e^{\frac{2\pi i}{n+1}}$ gives you f(z,n)=0. This is infimum.
