Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was given a homework question to calculate $f^{(n)}(0)$ where $f(z)=\frac{1}{1+z}$.

I now have a way to solve the question, but I don't understand why another way I tried gives a different and incorrect result.

This is what I did:

We know that if $f(z)$ is analytic in and on $C$ then $$\oint_{C}\frac{f(z)}{(z-z_{0})^{n+1}}\, dz=\frac{2\pi i}{n!}f^{(n)}(z_{0})$$

In particular take $z_{0}=0,f(z)=\frac{1}{1+z}$ to get $$f^{(n)}(0)=\frac{n!}{2\pi i}\oint_{C}\frac{1}{(1+z)z^{n+1}}\, dz$$

It remains to calculate $$\oint_{C}\frac{1}{(1+z)z^{n+1}}\, dz$$

where $C$ is any s.t $f(z)$ is analytic in and on $C$ , take $C$ to be a circle around the origin with radius $0.5$ (so that $z=-1$ is not in or on $C$).

Denote $g(z)=\frac{1}{(1+z)z^{n+1}}$ then $$g(\frac{1}{z})=\frac{z^{n+1}}{1+\frac{1}{z}}=\frac{z^{n+1}}{\frac{1+z}{z}}=\frac{z^{n+2}}{1+z}$$ and $$\frac{1}{z^{2}}g(\frac{1}{z})=\frac{z^{n}}{1+z}=z^{n}\frac{1}{1-(-z)}=z^{n}\sum_{k=0}^{\infty}(-z)^{k}$$

But this means the residue of $\frac{1}{z^2}g(\frac{1}{z})$ at $z=0$ is $0$ and so the integral evaluates to $0$.

But calculating $f^{(1)}$ we see that the result is incorrect.

Can someone please explain to me what I did wrong ?

share|cite|improve this question
Why are you evaluating $g(\frac{1}{z})$? – copper.hat Jan 12 '13 at 19:16
Why do you evaluate the residue of $g(1/z)/z^2$ and not the residue of $g(z)$??? – Fabian Jan 12 '13 at 19:17
By the way: the question is most easily solved by remembering that $1/(1+z) = \sum_{n} (-z)^n$. – Fabian Jan 12 '13 at 19:18
@copper.hat - I am using the following theorem: If a function $f$ is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour $C$, then $\oint_{C}\, f(z)\, dz=2\pi iRes_{z=0}\frac{1}{z^{2}}f(\frac{1}{z})$ – Belgi Jan 12 '13 at 19:20
@Belgi: $g$ has a singular point at $z=-1$, which is outside the circle. The result you are using depends on $g$ being analytic outside the contour. – copper.hat Jan 12 '13 at 19:33
up vote 3 down vote accepted

$g$ has a singular point at $z=−1$, which is outside the circle around the origin of radius $\frac{1}{2}$. The result you are using depends on $g$ being analytic outside the circle.

As an aside, the formulation of the result you were trying to use depends on some contour $C$, which, in my opinion, confuses the issue. I prefer the following formulation of the corresponding result (which is Exercise V.2.12 in Conway's, "Functions of one complex variable"): If $f$ is analytic except for isolated singularities at $a_1,...,a_m$, then $\text{Res }(f,\infty) = -\sum_{k=1}^m \text{Res }(f,a_k)$, where $\text{Res }(f,\infty)$ is defined as the residue of $z \mapsto -\frac{1}{z^2} f(\frac{1}{z})$ at $z=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.