Analytic function on a disc Let $\delta$ $>$ $0$, $k$ $\in$ $\mathbb{N}$ and $f$ is analytic on $B(0,2\delta)$
with $|f^{(k)}(0)|$ =$k!\delta^{-k}${sup$|f(z)|$:$|z|=\delta$}. Then show that $f(z)$ = $az^k$ for some $a$ $\in$ $\mathbb{C}$. My idea was to use the Cauchy Integral Formula and get a bound on the coefficients after performing the Taylor expansion about $0$, but since $\delta$ is fixed, this is not getting me anywhere.Thanks for any help in advance.
 A: Making a rigorous proof for this turned out to be way more involved than what I'd expected.

From Cauchy's integral formula we have that for all $n \in \mathbb N$
$$f^{(n)}(0) = \frac{n!}{2\pi i} \,\oint_{C(0;\delta)}\,\frac{f(z)}{z^{n+1}}\,dz.$$
Hence:
\begin{align}
\left|f^{(n)}(0)\right|
&=
\frac{n!}{2\pi} \,\left| \oint_{C(0;\delta)}\,\frac{f(z)}{z^{n+1}}\,dz\right|
\\&\leq
\frac{n!}{2\pi} \,\oint_{C(0;\delta)}\,\left|\frac{f(z)}{z^{n+1}}\right|\,|dz|\tag{1}
\\&=
\frac{n!}{2\pi} \,\oint_{C(0;\delta)}\,\frac{\left|f(z)\right|}{\delta^{n+1}}\,|dz|
\\&\leq
\frac{n!}{2\pi} \cdot \left(\sup_{|z| = \delta}|f(z)|\right)\cdot\frac{1}{ \delta^{n+1} }\cdot \oint_{C(0;\delta)}\,1\,|dz|
\tag{2}
\\&=
\frac{n!}{2\pi} \cdot \left(\sup_{|z| = \delta}|f(z)|\right) \cdot\frac1{\delta^{n+1}} \cdot 2\pi\delta
\\&=
\frac{n!}{\delta^n} \cdot \left(\sup_{|z| = \delta}|f(z)|\right).
\end{align}
Therefore, if $\left|f^{(k)}(0)\right| = \frac{k!}{\delta^k} \cdot \left(\sup_{|z| =  \delta}|f(z)|\right)$, we would need to have equality on each passage above when $n=k$.
Equality in passage $(2)$ would mean $|f|$ is constant on $C(0;\delta)$, with value $\frac1{k!}\,\delta^k\,\left|f^{(k)}(0)\right|$.
Equality in passage $(1)$ would imply
$$\left| \oint_{C(0;\delta)}\,\frac{f(z)}{z^{k+1}}\,dz\,\right|
=
\oint_{C(0;\delta)}\,\left|\frac{f(z)}{z^{k+1}}\right|\,|dz|.$$
We now apply the lemma (see the end), with $\gamma(t) = \delta e^{it}$. 
We have $\gamma'(t) = i \delta e^{it}$, so that $|\gamma'(t)| = \delta$ and hence
$$\frac{\overline{\gamma'(t)}}{|\gamma'(t)|}
=
\frac{-i \delta e^{-it}}{\delta}
=
\frac{-i \delta }{\delta e^{it}}
=
\frac{-i \delta }{\gamma(t)}.$$
Using the lemma with $z=\gamma(t)$, it follows that there is $\theta \in\mathbb R$ such that on $C(0;\delta)$
$$\frac{f(z)}{z^{k+1}}
=
e^{i\theta}\,\frac{-i \delta}{z} \cdot \frac{\frac{\delta^k}{k!}\,\left|f^{(k)}(0)\right|}{\delta^{k+1}}
\implies
f(z) = -ie^{i\theta}\, \frac{\left|f^{(k)}(0)\right|}{k!} \cdot z^k
.$$
Finally, with $a=-ie^{i\theta}\, \frac{1}{k!}\, \left|f^{(k)}(0)\right|$, the identity theorem guarantees that $f(z) = a\, z^{k}$ throughout $B(0;2\delta)$, which concludes our proof. $\square$


Lemma:
  Let $g$ be holomorphic on some open set $U\subset \mathbb C$ and $\Gamma \subset U$ be a rectifiable, continuously differentiable path.
  If $g$ satisfies
$$
\left|\int_{\Gamma}\,g(z)\,dz\,\right|
=
\int_{\Gamma}\,\left|g(z)\right|\,|dz|
$$
and $\gamma:[a,b]\longrightarrow \Gamma$ is a regular, continuously differentiable parametrization of $\Gamma$, then there is some $\theta \in \mathbb R$ such that
$$
g\big(\gamma(t)\big) = e^{i\theta}\, \frac{\overline{\gamma'(t)}}{\left|\gamma'(t)\right|} \cdot \left|g\big(\gamma(t)\big)\right|
$$
for all $t\in[a,b]$.
  In particular, if $\gamma$ is unit speed then $g\big(\gamma(t)\big) = e^{i\theta}\, \overline{\gamma'(t)} \cdot \left|g\big(\gamma(t)\big)\right|$.

Proof:
Let $\gamma:[a,b]\longrightarrow \Gamma$ be a regular, continuously differentiable parametrization of $\Gamma$.
Let $\alpha =$ $\int_{\Gamma}\,g(z)\,dz \in \mathbb C$ and write it in polar coordinates as $\alpha = re^{i\theta}$.
We then have $|\alpha| = \alpha \,e^{-i\theta}$ and hence
\begin{align}
\left|\int_{\Gamma}\,g(z)\,dz\,\right|
&=
\int_{\Gamma}\,e^{-i\theta}\,g(z)\,dz
\\&=
\int_a^b \, e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t) \, dt
\\&=
\int_a^b \, \Re\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right) \, dt
+i\cdot
\underbrace{
\int_a^b \, \Im\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right) \, dt
}_{0}\,
\\&\leq
\int_a^b \, \left|e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right| \, dt
\\&=
\int_a^b \, \left| g\left(\gamma(t)\right) \cdot \gamma'(t)\right| \, dt
=
\int_{\Gamma}\,\left|g(z)\right|\,|dz|
.\end{align}
If $\left|\int_{\Gamma}\,g(z)\,dz\,\right| = \int_{\Gamma}\,\left|g(z)\right|\,|dz|$, we'd have
$$
\int_a^b \,
\left|g\left(\gamma(t)\right) \cdot \gamma'(t)\right|
-
\Re\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right)
\, dt
= 0
$$
This is a real integral, and the integrand is $\geq 0$, so the integrand must vanish throughout the domain of integration.
Therefore we have
$$\left|g\left(\gamma(t)\right) \cdot \gamma'(t)\right|
=
\Re\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right)$$
for all $t\in [a,b]$.
It follows that $\Im\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right) = 0$ for all $t\in [a,b]$, and hence
\begin{align}
&e^{-i\theta} g\left(\gamma(t)\right) \cdot \gamma'(t)\
=
\Re\left(e^{-i\theta} \, g\left(\gamma(t)\right) \cdot \gamma'(t)\right)
=
\left|g\left(\gamma(t)\right) \cdot \gamma'(t)\right|
\\\implies\,&
g\left(\gamma(t)\right) = e^{i\theta}\, \frac{|\gamma'(t)|}{\gamma'(t)} \cdot \left|g\left(\gamma(t)\right)\right|
\\\implies\,&\vphantom{a^{b^{b^{b^{b^{b^{b^{b^{b^{b}}}}}}}}}}
g\left(\gamma(t)\right) = e^{i\theta}\, \frac{\overline{\gamma'(t)}}{\left|\gamma'(t)\right|} \cdot \left|g\left(\gamma(t)\right)\right|
\end{align}
for all $t\in[a,b]$, as we sought to prove. $\square$
As a bonus, we showed that the $e^{i\theta}$ in the lemma is given by
$$e^{i\theta} =
\frac{
\int_{\Gamma}\,g(z)\,dz
}{
\left| \int_{\Gamma}\,g(z)\,dz\right|
}.$$
A: Lemma: Suppose $g:[a,b]\to \mathbb C$ is continuous. Let $M= \sup_{[a,b]}|g|.$ If
$$|\int_a^b g(t)\,dt\,| = M (b-a),$$
then $g$ is constant on $[a,b].$ 
The proof of the lemma is a nice exercise. Ask questions if you like.
To our problem: Let $M=\sup_{t\in [0,2\pi]} |f(\delta e^{it})|.$ By Cauchy's integral formula, we get, after the standard parameterization and simplification,
$$2\pi\left |\frac{ f^{(k)}(0)}{k!}\right| = \left |\int_0^{2\pi} \frac{f(\delta e^{it})}{(\delta e^{it})^k}\,dt \right| \le M\delta^{-k}\cdot 2\pi.$$
By hypothesis, the first and last expression are equal. Therefore the middle and last expression are equal. But notice that $M\delta^{-k}$ is the $\sup$ of the modulus of the integrand on $[0,2\pi].$ Hence by the lemma, $f(z)/z^k$ is a constant $c$ on the circle $|z|= \delta$ for some constant $c.$ I.e, $f(z)=cz^k$ on this circle. By the identity principle, $f(z) = cz^k$ everywhere. 
