Prove $|1 - (1+\frac{z-1}{n})^m| \leq 1$ where $|z|\leq 1$ and $mHow to prove
$$\Big|1 - \big(1+\frac{z-1}{n}\big)^m\Big| \leq 1$$
where $z\in\mathbb{C}$, $|z|\leq 1$ and $m,n\in\mathbb{N}$ and $m<n$.
I have run numerical experiments and believe the inequality is correct.

By Maximum modulus principle, the maximum of $\displaystyle |1-(1+\frac{z-1}{n})^m|$ cannot be reached when $|z| < 1$. So we can suppose $|z|=1$.
Let $z=e^{i\theta}$, and $\displaystyle w = 1 + \frac{z-1}{n} = \rho e^{i \varphi}$. Then:
$$
\begin{align}
|1-(1+\frac{z-1}{n})^m | &= |1-w^m| \\
&=|1-w|\cdot|1+w+\cdots+w^{m-1}| \\
&\leq |1-w|\cdot (1+|w|+\cdots+|w|^{m-1}) \\
&\leq |1-w|\cdot (1+|w|+\cdots+|w|^{n-2})  \\
&=  \sqrt{1 -2\cos \varphi \cdot \rho+\rho^2} \cdot(1+\rho+\cdots \rho^{n-2})
\end{align}
$$
Since
$
\displaystyle |\rho e^{i\varphi}-(1-\frac{1}{n})|= |\frac{e^{i\theta}}{n}|=\frac{1}{n}
$,
$$
\rho^2+(1-\frac{1}{n})^2-2\rho\cdot (1-\frac{1}{n})\cos\varphi=\frac{1}{n^2}
$$
. Hence
$$
2\cos \varphi \cdot \rho=\frac{\rho^2+(1-\frac{2}{n})}{1-\frac{1}{n}}
$$
. Hence
$$
\begin{align}
\sqrt{1 -2\cos \varphi \cdot \rho+\rho^2} &= \sqrt{1+\rho^2-\frac{\rho^2+(1-\frac{2}{n})}{1-\frac{1}{n}}} \\
&=\sqrt{\frac{1-\rho^2}{n-1}}
\end{align}
$$
So if we can prove
$$
(1-\rho^2)(1+\rho+\cdots+\rho^{n-2})^2 \leq n-1
$$, we can prove the original statement. Here we know that $\displaystyle \rho \in [1-\frac{2}{n}, 1]$.
Let $t^2 = 1-\rho$, then
$$
\begin{align}
(1-\rho^2)(1+\rho+\cdots+\rho^{n-2})^2 &= (2-t^2)t^2 \frac{(1-(1-t^2)^{n-1})^2}{t^4} \\
&\leq 2\frac{(1-(1-t^2)^{n-1})^2}{t^2}
\end{align}
$$
Hence if we can prove that
$$
1 - \sqrt{\frac{n}{2}}t \leq (1-t^2)^{n}
$$, we can prove the original problem.
But I cannot prove this statement. Plotting some graph for small $n$, I believe it's true for all $n \geq 4$ and $t \in [0, 0]$. I asked a new question about this statement: Prove that $1 - \sqrt{\frac{n}{2}}\cdot t \leq (1-t^2)^n$ for $n \geq 4$ and $t \in [0, 1]$
