Does the logarithm inequality extend to the complex plane? For estimates, the inequality $\log(y)\le y-1,$ $y>0$ is often helpful. Is there any sort of upper bound for the logarithm function in the complex plane? Specifically, $|\log(z)|\le$ something for all $z \in \mathbb C$
Perhaps this would work?:
$\log(z)\le\sqrt{\log^2|z|+\arg(z)^2}$
 A: For the principal branch $\log z = \log|z| + i\arg z\;$ you get the somewhat trivial inequality
$$|\log z| = |\log|z| + i\arg z| \le |\log|z|| + \pi$$
Another restricted one is from http://dlmf.nist.gov/4.5
$$|\log (1+z)| \le -\log(1-|z|), \quad |z|<1$$
A: 
GLOBAL BOUND:

If one is seeking a global bound, then we have on the principal branch $\log(z)=\text{Log}(z)=\log(|z|)+\text{Arg}(z)$, where $-\pi<\text{Arg}(z)\le \pi$, and for $|z|> 0$
$$\begin{align}
|\text{Log}(z)|&=|\log(|z|)+i\text{Arg}(z)|\\\\
&\le \sqrt{\log^2(|z|)+\pi^2}\\\\
\end{align}$$
For $|z|\ge 1$, we have
$$\begin{align}
|\text{Log}(z)|\le \sqrt{(|z-1|)^2+\pi^2}\\\\
\end{align}$$
while for $0<|z|<1$, we have
$$\begin{align}
|\text{Log}(z)|\le \sqrt{\left|\frac{z-1}{z}\right|^2+\pi^2}\\\\
\end{align}$$


A USEFUL LOCAL BOUND:

If we restrict $|z|$ such that $|z-1|\le \rho$ for $\rho \in (0,1)$, then we can find a useful upper bound.  We write $\text{Log}(z)=\int_0^1\frac{z-1}{1+(z-1)t}\,dt$ such that  
$$\begin{align}
|\text{Log}(z)|&=\left|\int_0^1\frac{z-1}{1+(z-1)t}\,dt\right|\\\\
&\le \int_0^1\frac{|z-1|}{|1+(z-1)t|}\,dt\\\\
&\le \int_0^1\frac{|z-1|}{|1-|z-1|t|}\,dt\\\\
&\le \frac{|z-1|}{1-\rho} \tag 1
\end{align}$$
For example, if we take $\rho =1/2$, then $|\text{Log}(z)|\le 2|z-1|$ for $|z-1|<1/2$.


APPLICATION:

One application of the inequality $(1)$ is to prove that the infinite product representation for the sinc function, given by $\frac{\sin(\pi z)}{\pi z}=\prod_{n=1}^\infty\left(1-\frac{z^2}{n^2}\right)$ converges uniformly on compact sets.  
Equipped with $(1)$, we have for $|z|<B$ and $N>B/\sqrt\rho$
$$\begin{align}
\left|\log\left(\prod_{n=N}^\infty\left(1-\frac{z^2}{n^2}\right)\right)\right|&=\left|\sum_{n=N}^\infty \log\left(1-\frac{z^2}{n^2}\right)\right|\\\\
&\le \frac1{1-\rho }\sum_{n=N}^\infty \frac{|z^2|}{n^2}\\\\
&\le \frac{B^2}{1-\rho}\sum_{n=N}^\infty \frac{1}{n^2}
\end{align}$$
For any $\epsilon>0$, there exists a number $N'$ such that whenever $N>N'$, $\sum_{n=N}^\infty \frac{1}{n^2}<\epsilon(1-\rho)/B^2$.  And hence, the convergence of the infinite product representation of the sinc function is uniform.
A: Log denotes the principal branch of the complex logarithm
For $z \neq 0$, we have:
$Log(z) \overset{\Delta}{=} \log(|z|) + iarg(z)$
So:
$\left| Log(z) \right| \leq |\log(|z|)|+|arg(z)| \leq |z|+1+\pi$
