# Complex Analysis - Cauchy's Inequality problem

I need help with the following question.

Question:

Let $f$ be an entire function and $M$ be a constant such that $$|(f(z))| \le M|z|^\frac{5}{4}$$ for all $z\in C$. Show that we can find a constant $a \in C$ such that $f(z)=az$ for all $z \in C$. ($C$ denotes the complex numbers).

My attempt so far:

Cauchy's inequality says: $$|f'(z_0)| \le \frac{max_{|z-z_0|=R}|f(z)|}{R}$$ We know $|(f(z))| \le M|z|^\frac{5}{4}$. We also know $|z| = |z+z_0 - z_0|$, and by the triangle inequality we get $|z|=|z+z_0 -z_0| \le |z-z_0|+|z_0| = R+|z_0|$. This implies $|z|^\frac{5}{4} \le (R+|z_0|)^\frac{5}{4}$, which gives: $$|(f(z))| \le M(R+|z_0|)^\frac{5}{4}$$ Plugging this into Cauchy's Inequality gives $$|f'(z_0)| \le \frac{M(R+|z_0|)^\frac{5}{4}}{R}$$ I don't really know where to go from here to complete the proof. I've been told to use Liouville's Theorem, but I'm not sure how. My initial thought was to let $R$ go to infinity, (so our disc covers the entire complex plane), but then the right hand side of the above inequality would just go to $0$, and we would end up with $f(z) = constant$. Any help would be appreciated.

The inequality implies $f(0) = 0$. So $f(z) = zg(z)$ for some entire function $g$. For all $r > 0$, $$|g^{(n)}(0)| \le r^{-n} \max_{|z| = r}|g(z)| \le Mr^{\frac{1}{4} - n}$$
The last expression tends to $0$ as $r\to \infty$ as long as $n \ge 1$. Thus $g^{(n)}(0) = 0$ for all $n\ge 1$, which implies $g$ is constant. Thus $f(z) = az$ for some constant $a\in \Bbb C$.
• Sorry I'm still confused. Why does the inequality imply $f(0)=0$? And how does this give $f(z)=zg(z)$? – Arron Apr 5 '15 at 21:08
• Setting $z = 0$ in the inequality, we get $|f(0)| \le M|0|^{5/4}$, or $|f(0)| \le 0$. Therefore $f(0) = 0$. Since $f$ is entire, there are constants $a_n$ such that $f(z) = \sum_{n = 0}^\infty a_n z^n$ for all $z\in \Bbb C$. Since $f(0) = 0$, $a_0 = 0$, thus $f(z) = zg(z)$ where $g(z) = \sum_{n = 0}^\infty a_{n+1}z^n$. – kobe Apr 5 '15 at 21:11
• Would it be correct to say - setting $z_0 =0$, you get $|f'(0)| \le \frac{M|R|^\frac{5}{4}}{R}$. Then as R goes to infinity, you get $|f'(0)| \le 0$ which implies $|f(0)|=0$? – Arron Apr 5 '15 at 21:17
• No. Note $\frac{M|R|^{5/4}}{R} = MR^{1/4}$, and $R^{1/4}\to \infty$ as $R\to \infty$. – kobe Apr 5 '15 at 21:22