# Does the logarithm inequality extend to the complex plane? For estimates, the inequality log(y)<=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)|<= something for all z in bbbC Perhaps this would work?: log(z)<=\sqrt(\log^2|z|+arg(z)^2)

Does the logarithm inequality extend to the complex plane?
For estimates, the inequality $\mathrm{log}\left(y\right)\le y-1,$$y>0$ is often helpful. Is there any sort of upper bound for the logarithm function in the complex plane? Specifically, $|\mathrm{log}\left(z\right)|\le$ something for all $z\in \mathbb{C}$
Perhaps this would work?: $\mathrm{log}\left(z\right)\le \sqrt{{\mathrm{log}}^{2}|z|+\mathrm{arg}\left(z{\right)}^{2}}$
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Genesis Rosario
For the principal branch $\mathrm{log}z=\mathrm{log}|z|+i\mathrm{arg}z\phantom{\rule{thickmathspace}{0ex}}$ you get the somewhat trivial inequality
$|\mathrm{log}z|=|\mathrm{log}|z|+i\mathrm{arg}z|\le |\mathrm{log}|z||+\pi$
Another restricted one
$|\mathrm{log}\left(1+z\right)|\le -\mathrm{log}\left(1-|z|\right),\phantom{\rule{1em}{0ex}}|z|<1$
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Melina Barber
GLOBAL BOUND:
If one is seeking a global bound, then we have on the principal branch $\mathrm{log}\left(z\right)=\text{Log}\left(z\right)=\mathrm{log}\left(|z|\right)+\text{Arg}\left(z\right)$, where $-\pi <\text{Arg}\left(z\right)\le \pi$, and for $|z|>0$
$\begin{array}{rl}|\text{Log}\left(z\right)|& =|\mathrm{log}\left(|z|\right)+i\text{Arg}\left(z\right)|\\ \\ & \le \sqrt{{\mathrm{log}}^{2}\left(|z|\right)+{\pi }^{2}}\\ \end{array}$
For $|z|\ge 1$, we have
$\begin{array}{r}|\text{Log}\left(z\right)|\le \sqrt{\left(|z-1|{\right)}^{2}+{\pi }^{2}}\\ \end{array}$
while for $0<|z|<1$, we have
$\begin{array}{r}|\text{Log}\left(z\right)|\le \sqrt{{|\frac{z-1}{z}|}^{2}+{\pi }^{2}}\\ \end{array}$
A USEFUL LOCAL BOUND:
If we restrict $|z|$ such that $|z-1|\le \rho$ for $\rho \in \left(0,1\right)$, then we can find a useful upper bound. We write $\text{Log}\left(z\right)={\int }_{0}^{1}\frac{z-1}{1+\left(z-1\right)t}\phantom{\rule{thinmathspace}{0ex}}dt$ such that
$\begin{array}{rl}|\text{Log}\left(z\right)|& =|{\int }_{0}^{1}\frac{z-1}{1+\left(z-1\right)t}\phantom{\rule{thinmathspace}{0ex}}dt|\\ \\ & \le {\int }_{0}^{1}\frac{|z-1|}{|1+\left(z-1\right)t|}\phantom{\rule{thinmathspace}{0ex}}dt\\ \\ & \le {\int }_{0}^{1}\frac{|z-1|}{|1-|z-1|t|}\phantom{\rule{thinmathspace}{0ex}}dt\\ \\ \text{(1)}& & \le \frac{|z-1|}{1-\rho }\end{array}$
For example, if we take $\rho =1/2$, then $|\text{Log}\left(z\right)|\le 2|z-1|$ for $|z-1|<1/2$
APPLICATION:
One application of the inequality $\left(1\right)$ is to prove that the infinite product representation for the sinc function, given by $\frac{\mathrm{sin}\left(\pi z\right)}{\pi z}=\prod _{n=1}^{\mathrm{\infty }}\left(1-\frac{{z}^{2}}{{n}^{2}}\right)$ converges uniformly on compact sets.
Equipped with $\left(1\right)$, we have for $|z| and $N>B/\sqrt{\rho }$
$\begin{array}{rl}|\mathrm{log}\left(\prod _{n=N}^{\mathrm{\infty }}\left(1-\frac{{z}^{2}}{{n}^{2}}\right)\right)|& =|\sum _{n=N}^{\mathrm{\infty }}\mathrm{log}\left(1-\frac{{z}^{2}}{{n}^{2}}\right)|\\ \\ & \le \frac{1}{1-\rho }\sum _{n=N}^{\mathrm{\infty }}\frac{|{z}^{2}|}{{n}^{2}}\\ \\ & \le \frac{{B}^{2}}{1-\rho }\sum _{n=N}^{\mathrm{\infty }}\frac{1}{{n}^{2}}\end{array}$
For any $ϵ>0$, there exists a number ${N}^{\prime }$ such that whenever $N>{N}^{\prime }$, $\sum _{n=N}^{\mathrm{\infty }}\frac{1}{{n}^{2}}<ϵ\left(1-\rho \right)/{B}^{2}$. And hence, the convergence of the infinite product representation of the sinc function is uniform.