# A complex log inequality

Standard

Let $z$ be a complex number. The power series expansion of $\text{log}(1+z)$ about $z_0=0$ is given by

$\displaystyle{\text{log}(1+z) = \sum_{n=1}^\infty (-1)^{n-1}\frac{z^n}{n} = z-\frac{z^2}{2} + \ldots}$

which has radius of convergence 1. If $|z|<1$ then

$\displaystyle{\left|1-\frac{\log(1+z)}{z}\right|\leq \frac{|z|}{2(1-|z|)}}$

If we further assume $|z|<1/2$ then

$\displaystyle{\left|1-\frac{\log(1+z)}{z}\right|\leq \frac{1}{2}}$

This gives,

$\displaystyle{\frac{|z|}{2}\leq |\log(1+z)|\leq \frac{3|z|}{2}, \quad |z|<\frac{1}{2}}$

I just wanted to see how this inequality will appear graphically, so here are the plots made using SageMath 7.5.1 (by fixing the real part of z to zero and varying the imaginary part till 1/2)

i = CDF.0; p1 = plot(lambda t: abs(log(1+t*i)), 0, 0.5, rgbcolor=(0.8,0,0),legend_label=’ $|log(1+z)|$’, thickness=2); p2 = plot(lambda t: abs((t*i)/2), 0, 0.5, rgbcolor=(0,0.8,0), legend_label=’$|z|/2$’, thickness=2); p3 = plot(lambda t: abs(3*(t*i)/2), 0, 0.5, rgbcolor=(0,0,0.8), legend_label=’ $3|z|/2$’, thickness=2); p1+p2+p3

I tried to get a graph where this inequlaity fails (i.e. the plots intersect), but failed to do so.