I Bounding modulus of complex logarithm times complex power function

[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.

 

[Office_Shredder]

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

 

[psie]

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

 

[Office_Shredder]

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 $z$

 

[pasmith]

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}<br /> |z^\lambda| &amp;= |e^{\Re(\lambda \ln z)}| \\<br /> &amp;= |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).