Bounding modulus of complex logarithm times complex power function

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• psie
psie
TL;DR Summary
In my textbook (Ordinary Differential Equations by Andersson and Böiers), they claim that ##|(\log(z))^jz^\lambda|## can be bounded above by ##c|z|^l## for some integer ##l## and constant ##c##. I have a hard time confirming this.
It is claimed that the modulus of ##(\log(z))^jz^\lambda##, where ##j## is a positive integer (or ##0##) and ##\lambda## a complex number, can be bounded above by ##c|z|^l## for some integer ##l## and constant ##c##. Assume we are on the branch ##0\leq \mathrm{arg}(z)<2\pi## (yes, ##0## included; hence a discontinuous logarithm). Anyway, here's what I've tried so far.

Let ##\lambda_1## and ##\lambda_2## be the real and imaginary part of ##\lambda## respectively. By definition, ##\log(z)=\ln(|z|)+i\mathrm{arg}(z)##. Then $$|(\log(z))^j|=|\log(z)|^j\leq\left(\sqrt{\ln(|z|)^2+4\pi^2}\right)^j,$$ and $$|z^\lambda|=|e^{\lambda\log(z)}|=e^{\lambda_1\ln(|z|)}e^{-\lambda_2\arg(z)}=|z|^{\lambda_1}e^{-\lambda_2\arg(z)}.$$

Then someone has pointed to the limit ##\lim _{x\to \infty }\frac{(\ln x)^r}{x^k}=0## for ##r,k>0##, yet I don't see how we can write my simplification as this limit, if I have understood things right. Maybe there's another approach. Grateful for any help.

Last edited:
Are you only interested in large values of z or is this supposed to be true near the origin as well?

Office_Shredder said:
Are you only interested in large values of z or is this supposed to be true near the origin as well?
I forgot to mention, it's supposed to be true near the origin, so for ##|z|>0## small. I have found a solution elsewhere and I think there's no harm in posting it here. Write ##\lambda=x+iy##. So

\begin{align} |(\log z)^j| &=|\log|z|+i\arg z|^j \nonumber \\ &\le\big(\big|\log|z|\big|+2\pi\big)^j \nonumber \\ &\le\left(\frac{1}{|z|}+2\pi\right)^j \nonumber \\ &\le\left(\frac{2}{|z|}\right)^j \nonumber \end{align}

The first inequality is the triangle inequality, second follows from ##ye^{-y}<1## when ##y=-\log|z|>0## is large and the third one is ##2\pi<\frac{1}{|z|}## for ##|z|## small. Moreover,

\begin{align} |z^\lambda| &=|\exp(\lambda\log z)| \nonumber \\ &=\exp\big(\operatorname{Re}(\lambda\log z)\big) \nonumber \\ &=\exp\big(x\log|z|-y\arg z\big) \nonumber \\ &=|z|^xe^{-y\arg z} \nonumber \\ &\le|z|^{\lfloor x\rfloor}e^{-y\arg z} \nonumber \\ &\le|z|^{\lfloor x\rfloor}e^{\max\{0,-2\pi y\}}. \nonumber \end{align}

Thus, ##|(\log z)^jz^\lambda|\le 2^je^{\max\{0,-2\pi y\}}|z|^{\lfloor x\rfloor-j}##.

I guess we can simplify this. If ##|\log(z)| \leq c|z|^n## then ##|\log(z)^j z^\lambda| \leq c^j |z|^{(nj+\lambda)}##. So it suffices to just consider ##\log(z)##.

Ignoring the discontinuity with the branch cut, we can try to just apply l'hospital's rule.

##\lim_{z\to 0} \log(z)/z^n## (with n negative, so you get an infinity over infinity) Taking one derivative gives ##(1/z)/( (nz^{n-1}) = 1/(nz^n)##. So even with ##n=-1## we get that this limit is zero, and hence ##|\log(z)| \leq |z^{-1}|## for small enough ##z##

psie
Office_Shredder said:
I guess we can simplify this. If ##|\log(z)| \leq c|z|^n## then ##|\log(z)^j z^\lambda| \leq c^j |z|^{(nj+\lambda)}##.

A little more work to do, since $\lambda$ is complex. $$\begin{split} |z^\lambda| &= |e^{\Re(\lambda \ln z)}| \\ &= |z|^{\Re(\lambda)}e^{-\Im(\lambda)\arg z} \end{split}$$ and having chosen your branch you can remove the dependence on $\arg z$ by maximising $-\Im(\lambda) \arg(z)$.

Office_Shredder

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