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- 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.

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.

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