Does the logarithm inequality extend to the complex plane?

Question

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?: $\log(z)\le\sqrt{\log^2|z|+\arg(z)^2}$

Answer 1

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

$$|\log (1+z)| \le -\log(1-|z|), \quad |z|<1$$

Answer 2

Log denotes the principal branch of the complex logarithm For $z \neq 0$, we have: $Log(z) \overset{\Delta}{=} \log(|z|) + iarg(z)$

So: $\left| Log(z) \right| \leq |\log(|z|)|+|arg(z)| \leq |z|+1+\pi$

Answer 3

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

$$\begin{align} |\text{Log}(z)|\le \sqrt{\left|\frac{z-1}{z}\right|^2+\pi^2}\\\\ \end{align}$$

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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 $|z-1|<1/2$.

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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.