$ |1+z+z^2 +...+z^n| \geq |z|^n $ is valid

Question

Let $z=a+ib$ , where $a>0$ & $b $ are integers, then $ |1+z+z^2 +...+z^n| \geq |z|^n $ is always holds...

I tried to prove the because this seems to be true... But in vain

Answer 1

The inequality is true. For $n=0,1$ the inequality is clear, hence let $n \geq 2$. The case $z=1$ is also clear, so let $z \neq 1$.

Note that $1+z+\dots + z^n = \frac{z^{n+1}-1}{z-1}$. So the given inequality is equivalent to $$ \left|z - \frac{1}{z^n} \right| \geq |z-1| \iff \left|z - \frac{1}{z^n} \right|^2 \geq |z-1|^2.$$ Now adhere to $|z|^2 = z \overline{z}$. So equivalently we have to prove $$ z \overline{z} - \left(\frac{z}{\overline{z}^n} + \frac{\overline{z}}{z^n} \right) + \frac{1}{z^n \overline{z}^n} \geq z\overline{z} - (z + \overline{z}) + 1.$$ This is equivalent to $$2 \operatorname{Re}(z) + \frac{1 - \left(z^{n+1} + \overline{z}^{n+1} \right)}{|z|^{2n}} \geq 1 \iff 2 \operatorname{Re}(z) + \frac{1 - 2 \operatorname{Re}\left(z^{n+1} \right)}{|z|^{2n}} \geq 1. $$ But the estimation $$ 2 \operatorname{Re}(z) + \frac{1 - 2 \operatorname{Re}\left(z^{n+1} \right)}{|z|^{2n}} \geq 2 \operatorname{Re}(z) + \frac{1}{|z|^{2n}} - \frac{2}{|z|^{n-1}} \overset{?}{\geq} 1 $$ almost finishes. Note that $\operatorname{Re}(z) \geq 1$ by assumption. As $z \neq 1$ by assumption, $\frac{2}{|z|^{n-1}} \leq 1$, so again, we're done.

Our estimations were somewhat crude, so this inequality can be strengthened considerably with a more detailed analysis.