Why is $\log(f(z))$ entire if $f(z)\ne0$ and $f$ is entire? 
Let $f:\mathbb{C}\to\mathbb{C}\setminus\{0\}$ be an entire function. Why is $\log(f(z))$ entire?

I don't understand the answer because if we have log with any branch $B=\{Re^{i\theta}:R\geq0\}$, (and assume we choose for example the principal branch, $\theta=0$), then it may be that $f(z_0)\in B$ for some $z_0\ne0$ and then $\log(f(z_0))$ is not defined.
 A: There is a certain subtlety regarding what $\log(f(z))$ means. It is not true that for any nonvanishing $f$ there is a branch of logarithm for which $\log(f(z))$ is defined everywhere (indeed, for e.g. $f(z)=e^z$, which is onto $\mathbb C\setminus\{0\}$, that would imply existence of a branch of logarithm defined everywhere apart from zero, which you probably know is impossible.
What is true, and what that statement implicitly means, is that for a nonvanishing $f$ there exists a function $g(z)$, which we can denote by $\log(f(z))$, such that $e^{g(z)}=f(z)$.
A: You have good reasons to find the question unclear. However, here is anothor way of stating it:

Let $f\colon\mathbb C\longrightarrow\mathbb C\setminus\{0\}$ be an entire function. Why is there an entire function $g\colon\mathbb C\longrightarrow\mathbb C$ such that$$(\forall z\in\mathbb C):e^{g(z)}=f(z)?$$

Just take a primitive $h$ of $\frac{f'}f$. It is not hard to prove that $\frac{e^h}f$ is constant. So, there is a $k\in\mathbb C$ such that $(\forall z\in\mathbb C):\frac{e^{h(z)}}{f(z)}=e^k$, and therefore $(\forall z\in\mathbb C):f(z)=e^{h(z)-k}$.
