- #1

LagrangeEuler

- 717

- 20

- Homework Statement
- Find the residue of function [tex]\ln\frac{\sqrt{z^2+1}}{z}[/tex] at ##z=0##.

- Relevant Equations
- Residue of the function is ##c_{-1}## in Laurent series

[tex]f(z)=\sum^{\infty}_{n=-\infty}f(z)(z-z_0)^n[/tex]

First of all I am not sure which type of singularity is ##z=0##?

[tex]\ln\frac{\sqrt{z^2+1}}{z}=\ln (1+\frac{1}{z^2})^{\frac{1}{2}}=\frac{1}{2}\ln (1+\frac{1}{z^2})=\frac{1}{2}\sum^{\infty}_{n=0}(-1)^{n}\frac{(\frac{1}{z^2})^{n+1}}{n+1}[/tex]

It looks like that ##Res[f(z),z=0]=0##

[tex]\ln\frac{\sqrt{z^2+1}}{z}=\ln (1+\frac{1}{z^2})^{\frac{1}{2}}=\frac{1}{2}\ln (1+\frac{1}{z^2})=\frac{1}{2}\sum^{\infty}_{n=0}(-1)^{n}\frac{(\frac{1}{z^2})^{n+1}}{n+1}[/tex]

It looks like that ##Res[f(z),z=0]=0##