If $f(z)$ is analytic in the disk $|z|<R$ and has expansion $$ f(z) = a_0 + a_1z + a_2z^2 + \cdots, $$ Cauchy's estimate states that $$ |a_n| \le \frac{M(r)}{r^n} $$ where $M(r) = \max_{|z| = r} |f(z)|$ for any $r<R$.
Now if for some $k$ and $0 < r_0 < R$, $$ |a_k| = \frac{M(r_0)}{r_0^k}, $$ prove that $$ f(z) = a_k z^k. $$
My first attempt was that the two equalities holds in $$ |a_k| = \left|\frac{1}{2\pi i}\int_{|z|=r_0} \frac{f(z)}{z^{k+1}}\,\mathrm{d}z\right| \le \frac{1}{r_0^k} \frac{1}{2\pi}\int_{|z|=r_0} |f(z)|\,\mathrm{d}z \le \frac{M(r_0)}{r_0^k}$$ only if $$ \arg\left(\frac{f(z)}{z^{k+1}}\,\mathrm{d}z\right) = \mathrm{const},\quad \mathrm{i.e.}\quad \arg\left(\frac{f(z)}{z^k}\right) = \mathrm{const}, $$ and $$ |f(z)| = M(r_0) $$ on $|z| = r_0$. Together they imply that $f(z) = c_kz^k$ on $|z| = r_0$, and thus in $|z| < R$.
However, this proof seems really strange: notations like $\arg\left(\frac{f(z)}{z^{k+1}}\,\mathrm{d}z\right)$ is unconventional, and conditions for the equality in the triangle inequality to hold in hardly invoked elsewhere in complex analysis. There may be a proof more elegant than this one. I came across a similar question Find all entire functions () such that |()|=1 for ||=1 [duplicate], but the strategy there seems inapplicable here.
The proof could be simpler if the equality $$ |a_k| = \frac{M(r)}{r^k} $$ holds for some $k$ and all $0 < r < R$. Which shows $f(z) = O(|z|^k)$ and thus $$ f(z) = a_kz^k + a_{k+1}z^{k+1} + \cdots. $$ Therefore $f(z)/z^k$ is analytic, $$ \max_{|z| = r} \left|\frac{f(z)}{z^k}\right| = \frac{M(r)}{r^k} = |a_k|,\quad \forall r<R. $$ By maximum modulus principle, $f(z)/z^k = \mathrm{const}$. Is there also a simpler proof to the case where $ |a_k| = \frac{M(r)}{r^k} $ holds for some $r=r_0$ only?
