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.
 A: 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 $t \in [0, 1]$
A: Based on @onriv's nice answer and user3750444's comment therein:
(Without using Maximum Modulus Principle)
Let $w = 1 + \frac{z - 1}{n}$
and $r = |w|$. We have
\begin{align*}
 \left|1 - w^m\right|
 &= |1 - w|\cdot |1 + w + w^2 + \cdots + w^{m-1}|\\
 &\le |1 - w|\cdot (1 + r + r^2 + \cdots + r^{m-1})\\
 &\le |1 - w|\cdot (1 + r + r^2 + \cdots + r^{n - 2}).
\end{align*}
We have $|w - 1 + 1/n|^2 = |z|^2/n^2 \le 1/n^2$
or
$$(1 - 1/n)^2 - (1 - 1/n)(w + \bar{w}) + |w|^2 \le 1/n^2$$
which results in
$$w + \bar{w} \ge \frac{(1 - 1/n)^2 + |w|^2 - 1/n^2}{1 - 1/n}.$$
Thus, we have
$$|1 - w|^2 = 1 + |w|^2 - (w + \bar{w})
\le 1 + |w|^2 - \frac{(1 - 1/n)^2 + |w|^2 - 1/n^2}{1 - 1/n}
= \frac{1 - |w|^2}{n - 1}.$$
Thus, we have
$$|1 - w^m| \le \frac{\sqrt{1  - r^2}}{\sqrt{n - 1}}(1 + r + r^2 + \cdots + r^{n - 2}).$$
We have $|w| = |1 - 1/n + z/n|
\le |1 - 1/n| + |z/n| \le 1$ and
$|w| = |1 - 1/n + z/n| \ge |1 - 1/n| - |z/n| \ge 1 - 2/n$. Thus, $|w| \in [1 - 2/n, 1]$.
It suffices to prove that, for all $n\ge 2$ and $r\in [1 - 2/n, 1]$,
$$\frac{\sqrt{1  - r^2}}{\sqrt{n - 1}}(1 + r + r^2 + \cdots + r^{n - 2})
\le 1. \tag{1}$$
If $r = 1$ or $n = 2, 3, 4$, it is easy to prove the inequality.
In the following, assume that $r < 1$ and $n \ge 5$.
The inequality is written as
$$\frac{\sqrt{1  - r^2}}{\sqrt{n - 1}}\cdot \frac{1 - r^{n-1}}{1 - r}
\le 1.$$
Using Bernoulli inequality $(1 - v)^s \ge 1 - sv$ for all $0 \le v < 1$ and $s\ge 1$, we have
$$r^{(n-1)/4} = [1 - (1 - r)]^{(n-1)/4}
\ge 1 - (1 - r)(n - 1)/4 \ge 0.$$
Also, we have
$$\frac{\sqrt{1  - r^2}}{\sqrt{n - 1}\, (1 - r)}
= \frac{1}{\sqrt{n - 1}}\sqrt{\frac{1 + r}{1 - r}}
\le \frac{1}{\sqrt{n - 1}}\sqrt{\frac{2}{1 - r}}.$$
It suffices to prove that
$$\sqrt{\frac{2}{(n - 1)(1 - r)}}
\left(1 - \left[1 - (1 - r)(n - 1)/4\right]^4\right)\le 1. $$
Letting $(n - 1)(1 - r)/2 = u^2$, it suffices to prove that, for all $u\in [0, \sqrt{1 - 1/n}]$,
$$\frac{1}{u}(1 - (1 - u^2/2)^4) \le 1$$
or
$$u^7 - 8u^5 + 24u^3 - 32u + 16 \ge 0$$
or
$$(1 - u)^3(-u^4 - 3u^3 + 2u^2 + 14u + 9) + 13(u - 19/26)^2 + 3/52 \ge 0$$
which is clearly true.
We are done.
