Daily Archives: December 2, 2017

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)

l1

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.

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