I'm learning about complex analysis and need to verify my work to this problem since my textbook does not provide any solution:
If $f(z)$ is entire and $|f(z)| \le \log(2+|z|)$ for every $z \in \Bbb C$ show that $f$ is constant.
Here's my attempt :
Suppose $z \in \Bbb C$ with $|z| \lt R$. By Cauchy's Integral Formula for Derivatives we have :
$$f'(z) = \frac{1}{2 \pi i} \int_{|\zeta|=R} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta \iff \left| f'(z) \right| = \frac{1}{2 \pi} \left| \int_{|\zeta|=R} \frac{f(\zeta)}{(\zeta - z)^2} \, d\zeta \right| \le$$
$$\le \frac{1}{2 \pi} \int_{|\zeta|=R} \frac{\left| f(\zeta) \right|}{\left| \zeta - z \right|^2} \, \left| d\zeta \right| \le \frac{1}{2 \pi} \int_{|\zeta|=R} \frac{\log(2+R)}{R^2} \, \left| d\zeta \right| \le$$
$$\le \frac{1}{2 \pi} \frac{\log(2+R)}{R^2} \int_{|\zeta|=R} \, \left| d\zeta \right| = \frac{1}{2 \pi} \frac{\log(2+R)}{R^2} 2 \pi R = \frac{log(2+R)}{R}$$
For $R \rightarrow \infty$ and $\forall z \in \Bbb C : \left| f'(z) \right| = 0 \implies f'(z) = 0 \implies f(z) = c$, where $c \in \Bbb C$ constant.
Is my work correct? Also, I noticed when studying the theory that when taking the absolute value of $f'(z)$ the complex constant $i$ is removed in Cauchy's integral Formula but I don't understand why. Can someone explain this?
The absolute value of $i$ is $1$ and so it is removed (using $|ab| = |a||b|$).
Regarding your work, note that in the second line, you replace $|\zeta - z|^2$ in the denominator of the integrand with $R^2$. If $z = 0$, then indeed $|\zeta - z|^2 = R^2$ but if $z \neq 0$, you need to estimate $|\zeta- z|^2$ and it is not true with your assumptions that $|\zeta - z| \geq R$. If you assume that $|z| < \frac{R}{2}$ (which you can, since you let $R \to \infty$ anyway), then using the reverse triangle inequality you have $|\zeta - z| \geq \frac{R}{2}$ and so you can continue your argument.