Show that $\operatorname{Res}\left(\frac{1}{f}, z_0\right)=\frac{1}{f'(z_0)}$

Question

I'm learning complex analysis, specifically (Laurent) series and residues, and need help to understand the solution to the following problem:

Let $f$ be holomorphic and non-constant on the open domain $\Omega$ and $f$ has a zero of order $1$ at $z_0 \in \Omega$. Show that

$$\operatorname{Res}\left(\frac{1}{f}, z_0\right)=\frac{1}{f'(z_0)}.$$

Solution:

Since $f$ is holomorphic in $\Omega$, by Taylor's theorem, $f$ can be expanded in a power series around $z_0$ of the form

$$f(z) = \sum_{n=0}^{\infty}a_n(z-z_0)^n, \,\, \text{where} \,\,a_n = \frac{f^{(n)}(z_0)}{n!}.$$

Now, since $f$ has a zero of order $1$ at $z_0$, then $a_0 = 0$ and therefore

$$f(z) = (z-z_0)\sum_{n=0}^{\infty}a_{n+1}(z-z_0)^n \,\, \text{with} \,\, a_1 \neq 0.$$

Now for the part that I don't understand:

Let $g(z) = \sum_{n=0}^{\infty}a_{n+1}(z-z_0)^n$. Note that $g(z_0) \neq 0$ (why?) and $g(z)$ holomorphic (where?). I don't understand why $g(z_0) \neq 0$. Is it because $f$ is non-constant by assumption? If I plug in $z_0$ in the power series representation for $g(z)$ I get $0$. Back to the solution.

Then, there exists $r > 0$ such that $g(z) \neq 0 \; \forall z \in D(z_0, r)$ (why?). I don't get this part either. Why does there exists a disc centered at $z_0$ of radius $r$ in which $g(z)$ is non-zero? Back to the solution.

Therefore, the function $1/g$ is analytic in $D(z_0, r)$. Now write

$$\frac{1}{f(z)} = \frac{1}{(z-z_0)g(z)} \implies \operatorname{Res}\left(\frac{1}{f}, z_0\right) = \frac{1}{g(z_0)}$$

where $g(z_0) = a_1$ (why?) $= \frac{f^{(1)}(z_0)}{1!}$. The solution then follows. Here, it is said that $g(z_0) = a_1$ but I don't understand why.

Answer 1

Here are some answers to your questions:

• "The reason $\mathbf{g(z_0)\not=0}$": The fact that $g(z_0) = a_1 \not= 0$ is due to the assumption that $f(z)$ has a zero of order $1$ at $z=z_0$. If $a_1 = 0$ then we would have $f(z) = (z-z_0)^2(a_2 + a_3(z-z_0)+\ldots)$ and the zero would be (at least) of order $2$ contradicting this assumption.

• "Where is $\mathbf{g(z)}$ holomorphic": Holomorphic means analytical everywhere (in it's domain). The domain here is $\mathbb{\Omega}$ (the same as $f$). We have two ways of writing $g$. First we have the power-series expanded around $z=z_0$ for which it follows by definition that $g$ is analytic at $z=z_0$. Secondly we also see that $g(z) = \frac{f(z)}{z-z_0}$ for $z\not=z_0$. Since both $f$ and $\frac{1}{z-z_0}$ is analytic for $z\not=z_0$ it follows that $g$ is also analytic here.

• "Why $\mathbf{g(z)\not =0}$ close enough to $\mathbf{z_0}$": The function $|g(z)|$ is continuous and since $|g(z_0)| \not = 0$ it follows by the definition of continuity that there exists a $\delta > 0$ such that $|g(z)| > \frac{|g(z_0)|}{2} > 0$ for all $|z-z_0| < \delta$.

• Why is $\mathbf{g(z_0) = a_1}$": This follows by taking $z=z_0$ in the definition of $g(z)$ and noting that all terms except the first, which is just $a_1$, vanish. As noted in the comments you need to be a bit careful about the first term since $(z-z_0)^0 = 1$.

Answer 2

Here is an alternative way to show a more general version.

Claim. $$\operatorname{Res}_{z_0}\left(\frac{f}{g}\right)=\frac{f(z_0)}{g'(z_0)}$$

Proof. If $f$ has a zero of any order in $z_0$ it can be removed (since $g$ has a zero of order 1) thus there is no principal part of the Laurent expansion and the residue is $0$. We can assume $f$ has no zero so we are looking at a simple pole. The expansion in a neighborhood of $z_0$ is

$$\frac{f(z)}{g(z)}=\frac{c_{-1}}{(z-z_0)}+\sum_{k=0}^\infty c_k(z-z_0)^k$$

where we see that $c_{-1}=\operatorname{Res}_{z_0}\left(\frac{f}{g}\right)$. Multiplying with $(z-z_0)$ yields

$$(z-z_0)\frac{f(z)}{g(z)}=\operatorname{Res}_{z_0}\left(\frac{f}{g}\right)+\underbrace{(z-z_0)}_{\to 0\,(z\to z_0)}\underbrace{\sum_{k=0}^\infty c_k(z-z_0)^k}_{\to c_0\,(z\to z_0)}$$

and therefore taking the limit we finally get

$$ \begin{align*} \operatorname{Res}_{z_0}\left(\frac{f}{g}\right) &= \lim_{z\to z_0}(z-z_0)\frac{f(z)}{g(z)} = \lim_{z\to z_0}\frac{f(z)}{\frac{g(z)}{(z-z_0)}} \overset{(1)}{=} \lim_{z\to z_0}\frac{f(z)}{\frac{g(z)-g(z_0)}{(z-z_0)}} \overset{(2)}{=} \frac{f(z_0)}{g'(z_0)}. \end{align*} $$

At $(1)$ we know that $g(z_0)=0$ and for $(2)$ we know $g'(z_0)\neq 0$ since $z_0$ is a simple zero.

Answer 3

Another solution is this...

We have $$I:=res\left(\dfrac{1}{f},z_0\right)=\dfrac{1}{2\pi i}\int_{|z-z_0|=\epsilon}\dfrac{dz}{f(z)},$$ where $\epsilon>0$ is very small, such that $f$ admits power series expansion in $U=\lbrace z\in \mathbb{C}:|z-z_0|<\epsilon\rbrace$, and in this set the unique zero of $f$ is $z_0$. Now, since $f$ has a zero of order $1$, we have(expand $f$ in power series) $f(z)=(z-z_0)g(z)$, where $g$ is an holomorphic function whithout zeros in $\lbrace z\in \mathbb{C}:|z-z_0|<\epsilon\rbrace$. (*If for contraddiction we assume that $g$ has another zero in $U$ we should have another zero of $f$ in $U$ ) At this point, by the Cauchy Integral Formula we obtain $$I=\dfrac{1}{g(z_0)},$$but $f(z)=(z-z_0)g(z)\Longrightarrow f'(z)=g(z)-(z-z_0)g'(z)\Longrightarrow f'(z_0)=g(z_0)$.

$*$ If $g(z_0)=0$ we have $g(z)=(z-z_0)h(z)$ and then $f(z)=(z-z_0)^2 h(z)$, this is a contraddiction!

Answer 4

If $f$ has a single zero at $z_0$ then the function $g(z):={f(z)-f(z_0)\over z-z_0}$ has a removable singularity at $z_0$. In fact, $g$ is analytic in a neighborhood $U$ of $z_0$, and $g(z_0)=f'(z_0)\ne0$. It follows that $h:={1\over g}$ is analytic in $U$, and that $${1\over f(z)}={h(z)\over z-z_0}\qquad(z\in\dot U)$$ has a first order pole at $z_0$ with residue $$h(z_0)={1\over g(z_0)}={1\over f'(z_0)}\ .$$