how can I give an elementary proof of Maximum Modulus Theorem for polynomials? how can I give an elementary proof of Maximum Modulus Theorem for polynomials?
I got proof, but not elementary.
This question in this book Conway.
 A: For polynomials, we can prove the maximum modulus principle elementarily, using only arithmetic (I count the binomial theorem as arithmetic) and basic properties of the modulus, conjugation, and the exponential function.
We prove it in the form

Let $p$ be a non-constant complex polynomial. Then $\lvert p(z)\rvert$ has no local maximum at any point.

Let $p(z) = \sum_{k=0}^n a_k z^k$ with $a_n \neq 0$ and $n\geqslant 1$ (so that $p$ is non-constant). Let $z_0\in \mathbb{C}$ arbitrary.
The crucial point is that we can expand $p$ in powers of $z-z_0$,
$$\begin{aligned}
p(z) &= \sum_{k=0}^n a_k (z_0 + (z-z_0))^k\\
&= \sum_{k=0}^n a_k \sum_{m=0}^k \binom{k}{m}z_0^{k-m}(z-z_0)^m\\
&= \sum_{m=0}^n \underbrace{\left(\sum_{k=m}^n a_k\binom{k}{m}z_0^{k-m}\right)}_{b_m}(z-z_0)^m,
\end{aligned}$$
and we have $b_n = a_n \neq 0$. Also, $b_0 = p(z_0)$. Let $r = \min \{m : 1\leqslant m \leqslant n, \, b_m\neq 0\}$. Then
$$p(z) = b_0 + (z-z_0)^rb_r + \sum_{m=r+1}^n b_m (z-z_0)^{m}.$$
Now choose an $\alpha\in \mathbb{R}$ such that $e^{i\alpha}b_r\overline{b_0}$ is a non-negative real number and define $\beta = \alpha/r$. [If $b_0 = 0$, any $\alpha$ will do, and if $b_0\neq 0$, then $\alpha$ is uniquely determined modulo multiples of $2\pi$.]
Then, for $0 < t < 1$ we have
\begin{align}
\lvert p(z_0 + te^{i\beta})\rvert
&= \left\lvert b_0 + t^re^{ir\beta}b_r + \sum_{m=r+1}^n b_m t^{m-r}e^{im\beta}\right\rvert\\
&\geqslant \left\lvert b_0 + t^re^{i\alpha}b_r\right\rvert - \sum_{m=r+1}^n \lvert b_m\rvert\cdot t^m\\
&= \lvert b_0\rvert + t^r\lvert b_r\rvert - \sum_{m=r+1}^n \lvert b_m\rvert\cdot t^m\\
&\geqslant \lvert b_0\rvert + t^r \lvert b_r\rvert - \sum_{m=r+1}^n \lvert b_m\rvert\cdot t^{r+1}\\
&= \lvert b_0\rvert + t^r\left(\lvert b_r\rvert - t\sum_{m=r+1}^n \lvert b_m\rvert\right),
\end{align}
where $\lvert b_0 + t^re^{i\alpha}b_r\rvert = \lvert b_0\rvert + t^r\lvert b_r\rvert$ by the choice of $\alpha$. Then constraining further to
$$0 < t < \frac{\lvert b_r\rvert}{1 + 2\sum_{m=r+1}^n \lvert b_m\rvert}$$
shows that for these $t$ we have
$$\lvert p(z_0 + te^{i\beta})\rvert \geqslant \lvert b_0\rvert + t^r\frac{\lvert b_r\rvert}{2} > \lvert b_0\rvert = \lvert p(z_0)\rvert,$$
hence $\lvert p(z)\rvert$ has no local maximum at $z_0$. Since $z_0$ was arbitrary, it follows that $\lvert p(z)\rvert$ has no local maximum at all.
It may be worth mentioning that choosing $\alpha$ so that $e^{i\alpha} b_r\overline{b_0} < 0$ leads with essentially the same computation to the conclusion that $\lvert p(z)\rvert$ can have local minima only at zeros of $p$, and in that way to an elementary proof of the fundamental theorem of algebra, that every non-constant complex polynomial has a zero in $\mathbb{C}$.
A: Let $p(z)$ be a polynomial and $a \in \mathbb{C} $ such that $|p(z)| \leq |p(a)| $. If $a$ is a zero, then the polynomial is trivial and hence constant. If not, then we can expand $$p(z) = \sum_{i=0}^n \frac{p^{(i)}(a)}{i!}(z-a)^i$$ and note that $|p(a)| \leq |p(z)| $ by a simple application of triangle inequality, telling us that $|p(z)|=k$. Now invoke Exercise III.3.17
