If Re(z) >0, then the value of |1+ z+ z^2...z^n| cannot be less than?

Question

Here are the options- a)$|z|^n - 1/|z|$ b) $|z|^n + 1/|z|$ c) $n|z|^n$ d) $n|z|^n + 1$

The answer is supposed to be $|z|^n - 1/|z|.$

How do I solve this question? I tried writing $1+ z+ z^2...z^n$ as $[z^{n+1} - 1]/(z-1),$ but I couldn't figure out what to do after that...

Answer 1

For $\Re(z)\geq 1/2$ and $z\neq 1$ we have $$|z-1|\leq |z|$$ and then $$\left|\sum_{j=0}^nz^j\right|=\left|\frac{z^{n+1}-1}{z-1}\right|\geq\left|\frac{z^{n+1}-1}z\right|\geq|z|^n-\frac1{|z|}$$

For $z=1$ it is a simple check.

I couldn't find a proof for $0<\Re(z)<1/2$ so far.

EDIT: It seems that it is false.
For $z=0.1+10i$ and $n=2$ we have $$|z^2+z+1|<99.7$$ $$|z|^2-1/|z|>99.9$$