In my textbook (Ordinary Differential Equations by Andersson and Böiers), it is claimed that the complex modulus of $(\log(z))^jz^\lambda$, where $j\geq0$ is a natural number 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 $0\leq \mathrm{arg}(z)<2\pi$ and moreover, that this bound should hold near the origin. I have a hard time confirming this. 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 $$|z^\lambda|=|e^{\lambda\log(z)}|=e^{\lambda_1\ln(|z|)}e^{-\lambda_2\arg(z)}=|z|^\lambda_1e^{-\lambda_2\arg(z)}.$$
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.
Let $\lambda=x+iy$. When $|z|>0$ is small, $$ \begin{aligned} |(\log z)^j| &=|\log|z|+i\arg z|^j\\ &\le\big(\big|\log|z|\big|+2\pi\big)^j\\ &\le\left(\frac{1}{|z|}+2\pi\right)^j\\ &\le\left(\frac{2}{|z|}\right)^j,\\ |z^\lambda| &=|\exp(\lambda\log z)|\\ &=\exp\big(\operatorname{Re}(\lambda\log z)\big)\\ &=\exp\big(x\log|z|-y\arg z\big)\\ &=|z|^xe^{-y\arg z}\\ &\le|z|^{\lfloor x\rfloor}e^{-y\arg z}\\ &\le|z|^{\lfloor x\rfloor}e^{\max\{0,-2\pi y\}}.\\ \end{aligned} $$ Therefore $|(\log z)^jz^\lambda|\le 2^je^{\max\{0,-2\pi y\}}|z|^{\lfloor x\rfloor-j}$.