Let $f,g$ be holomorphic around $0$. Now I've seen the formula $Res((f\circ g)g')=ord(g)Res(f)$. Now if $f$ has a simple pole or is complex differentiable at $0$ and $ord(g)>0$ (edit: where $ord(g)$ is the multiplicity of the zero of $g$ at $0$) this is easy to check using the Laurent series. Is this also true for more general $f$?
Residue formula
1
$\begingroup$
complex-analysis
-
0What is $ord(g)$? – 2017-01-12
-
0@gobucksmath edited – 2017-01-12
-
1Yes, it is true for more general $f$. Try to represent $\operatorname{Res} (f(g(z))g^\prime(z);0)$ in a form of integral. – 2017-01-13
-
0Ohh, so this formula is true because we can just see the left hand side as the integral around zero, but $ord(g)$ times, because if $\gamma$ goes around once, $g\circ\gamma$ goes around $ord(g)$ times, as we can write $g=(h)^{ord(g)}$ with $h$ bijective from a neighbourhood of $0$ to another neighburhood of $0$? – 2017-01-13
-
0But with this argument I only got that they are only equal up to the sign? How do I get that the sign is equal aswell? – 2017-01-13
-
1If $\gamma $ goes around once **counterclockwise** , $g\circ \gamma $ goes around $ord(g)$ times **counterclockwise**. Therefore we have $$ \operatorname{Res}((f\circ g)g^\prime)=\frac{1}{2\pi i}\int_\gamma f(g(z))g^\prime(z)dz=\frac{1}{2\pi i}\int_{g\circ\gamma} f(\zeta )d\zeta =ord(g)\operatorname{Res}(f).$$ No problem about the sign. – 2017-01-13
-
0How do we know that $g\circ\gamma$ goes counterclockwise (with the assumption that $\gamma$ goes counterclockwise)? Thanks alot for the help! I was so focused on the series expansion, I didn't even think of using the definition of the residue! – 2017-01-13
-
1It's one of the properties of analytic functions. It can be seen easily. If $$ g(z)=a_mz^m+a_{m+1}z^{m+1}+\cdots=a_mz^m\left(1+\frac{a_{m+1}}{a_m}z+\cdots\right),$$ then $$ |\arg g(z)-\arg\left(a_mz^m\right)|=\left|\arg \left(1+\frac{a_{m+1}}{a_m}z+\cdots\right)\right|<\varepsilon $$ for sufficiently small $z$. Hence if $z$ goes counterclockwise, $g(z)$ goes counterclockwise. – 2017-01-14
-
0Wow, this is a really nice and intuitive argument! I'm really thankful for your help! – 2017-01-14