# What is wrong in this solution for finding the $n-th$ derivative of $f(z)=\frac{1}{1+z}$ at the point $z_{0}=0$?

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 ?

-
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

$g$ 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 $z=0$.

-