Let $f: \mathbb{C} \to \mathbb{C}$ be analytic and suppose a $C \geq 0$ exists such that \begin{align*} |f(z)| \leq C(|z| + 1) \log(|z| + 1) \end{align*} for all $z \in \mathbb{C}$, where $\log: \mathbb{R} \to \mathbb{R}$ is the usual logarithm. Show that there exists an $a \in \mathbb{C}$, such that $f(z) = az \; (z \in \mathbb{C})$.
From the other exercises I've done, one usually uses the line integral definition of the coefficients $a_n$ to show that $a_n = 0$ for $n \geq 2$, so that it can be concluded that $f(z) = az + b$.
I initially approached this problem similarly, but the line integral seems to ugly to me to be calculated algebraically: \begin{align*} |a_n| = \frac{1}{2\pi}\left| \int_{|z| = \rho} \frac{f(z)}{z^{n + 1}} dz \right| \leq \frac{1}{2\pi}\int_{|z| = \rho} \frac{| f(z) |}{ |z|^{n + 1}} dz \leq \frac{1}{2\pi}\int_{|z| = \rho} \frac{|C (|z| + 1)\log(|z| + 1)|}{|z|^{n+1}} dz = \frac{1}{2\pi}\int_0^{2 \pi} \frac{C (\rho e^{it} + 1) \log(\rho e^{it} + 1)}{(\rho e^{it})^{n + 1}} i\rho e^{it} dt \end{align*}
After all this trouble I'm wondering whether this is the correct method of doing this, so if anyone could shine a light on where I went wrong or what the actual method is, I would be very thankful. Cheers!
You made a mistake in your computations. It should read: $$|a_n| \leq \int_0^{2\pi} \frac{C (\rho + 1) \log(\rho + 1)}{\rho^{n+1}} \rho \mathrm{d} t$$
Since for $\rho$ big enough, $\log(\rho+1) < \rho$, you have a majoration of the form $|a_n| \leq C' \rho^{3/2 - n}$ for some constant $C'$ and $\rho$ big enough. If $n \geq 2$, it follows that $a_n = 0$ (by letting $\rho \to \infty$).