Meromorphic functions and the Logarithm

Zeroes and poles

Definition 410.1(Singularity).

A point singularity of a function $f$ is a complex number $z_0$ such that $f$ is defined in a neighborhood of $z_0$ but not at the point $z_0$ itself. These are also called isolated singularities. We call $z_0\in \mathbb{C}$ a zero for the holomorphic function $f$ if $f(z_0)=0$.

Theorem 410.2(Stein & Shakarchi (2003) 3.1.1).

Suppose that $f$ is holomorphic in a connected open set $\Omega$, has a zero at a point $z_0\in \Omega$, and does not vanish identically in $\Omega$. Then there exists a neighborhood $U\subseteq \Omega$ of $z_0$, a non-vanishing holomorphic function $g$ on $U$, and a unique positive integer $m$ such that

$$f(z)=(z-z_0{)}^{m}g(z)$$

for all $z\in U$. We say $f$ has a zero of order $n$ at $z_0$. A zero of order $1$ is called simple.

Proof.

By Thm 377.12, we can assume that $f$ does not vanish on any neighborhood of $z_0$. Let $V\subseteq \Omega$ be an open disc centered at $z_0$. By Thm 377.8, $f$ has a power series expansion $f(z)={\sum }_{n=0}^{\infty }{a}_n(z-z_0{)}^n$ at $z_0$ which converges on $V$. Let $a_m$ be the first non-zero coefficient. Then, we can write

$$\begin{array}{rl}f(z)& =(z-z_0{)}^m(a_m+a_{m+1}(z-z_0)+\dots )\\ & =(z-z_0{)}^{m}g(z).\end{array}$$

By Thm 370.8, $g$ is holomorphic on $V$ since it is defined by a power series having the same radius of convergence as that of $f$. Since $g(z_0)=a_m\ne 0$ and holomorphic functions are continuous, there exists a neighborhood $U$ of $z_0$ on which $g$ is nonzero.

It remains to prove the uniqueness of $m$. Suppose there exists $k\ne m$ and non-vanishing holomorphic $h$ on $U$ such that $(z-z_0{)}^{k}h(z)=(z-z_0{)}^{m}g(z)$. WLOG, $k>m$. Then, we have

$$g(z)=(z-z_0{)}^{k-m}h(z)$$

away from $z_0$. Letting $z\to z_0$ gives $g(z_0)=0$, a contradiction.□

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Definition 410.3(Pole).

We say that a function $f$ defined and holomorphic in a deleted neighborhood of $z_0$ has a pole at $z_0$ if the function $1/f$, defined to be zero at $z_0$, is holomorphic in a (full) neighborhood of $z_0$.

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Theorem 410.4(Stein & Shakarchi (2003) 3.1.2).

If $f$ has a pole at $z_0\in \Omega$, then there exists a neighborhood $U$ of $z_0$, a non-vanishing function $h$ holomorphic on $U$ and a unique positive integer $n$ such that

$$f(z)=(z-z_0{)}^{-n}h(z)(z\in U\setminus \{z_0\}).$$

We say $z_0$ is a pole of order (or multiplicity) $n$. Order 1 poles are called simple.

Proof.

By Thm 2, $1/f(z)=(z-z_0{)}^{n}g(n)$, where $g$ is holomorphic an non-vanishing in a neighborhood of $z_0$. The result follows with $h(z)=1/g(z)$.□

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Corollary 410.5(Stein & Shakarchi (2003) 3.1.3).

If $f$ has a pole of order $n$ at $z_0$, then there exists a neighborhood $U$ of $z_0$ such that for $z\in U\setminus \{z_0\}$, we have

$$\begin{array}{r}f(z)=\underset{\text{principal part of }f\text{ at }{z}_0}{\underbrace{\frac{a_{-n}}{(z-z_0{)}^n}+\frac{a_{-n+1}}{(z-z_0{)}^{n-1}}+\cdots +\frac{\stackrel{\text{residue of }f\text{ at }{z}_0}{\overbrace{a_{-1}}}}{(z-z_0)}}}+G(z),\end{array}$$

where $G$ is a holomorphic function in all of $U$. We write ${\text{res}}_{z_0}f=a_{-1}$.

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Corollary 410.6(Stein & Shakarchi (2003) 3.1.4).

If $f$ has a pole of order $n$ at $z_0$, then

$${\text{res}}_{z_0}f=\underset{z\to z_0}{\lim }\frac{1}{(n-1)!}{\left(\frac{d}{dz}\right)}^{n-1}(z-z_0{)}^{n}f(z).$$

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The residue formula

Theorem 410.7(Theorem).

Suppose that $f$ is holomorphic in an open set containing a circle $C$ and its interior, except for a pole at $z_0$ inside $C$. Then

$${\int }_{C}f(z)dz=2\pi i{\text{res}}_{z_0}f.$$

Proof.

As in the proof of Thm 377.5, we may choose a keyhole contour that avoids the pole, and let the width of the corridor go to zero to see that

$${\int }_{C}f(z)dz={\int }_{C_{ϵ}}f(z)dz$$

where $C_{ϵ}$ is the small circle centered at the pole $z_0$ and of radius $ϵ$. Using Cor 5 on $f$, Cor 377.6 on the constant functions $a_{-i}$, and Thm 377.4,

$$\begin{array}{rl}{\int }_{C_{ϵ}}f(z)dz& =\sum _{i=2}^n\underset{=0\forall i}{\underbrace{{\int }_{C_{ϵ}}\frac{a_{-i}}{(z-z_0{)}^i}dz}}+\underset{=2\pi i{a}_{-1}}{\underbrace{{\int }_{C_{ϵ}}\frac{a_{-1}}{(z-z_0)}dz}}+\underset{=0}{\underbrace{{\int }_{C_{ϵ}}G(z)dz}}\\ & =2\pi i{\text{res}}_{z_0}f.\end{array}$$ □

This theorem can be generalized to the case of finitely many poles.

Corollary 410.8(The residue formula, Stein & Shakarchi (2003) 3.2.2).

Suppose that $f$ is holomorphic in an open set containing a circle $C$ and its interior, expect for poles at the points $z_1,\dots ,z_N$ inside $C$. Then

$${\int }_{C}f(z)dz=2\pi i\sum _{k=1}^N{\text{res}}_{z_k}f.$$

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Cor 8 can also be stated for toy contours.

Singularities and meromorphic functions

Definition 410.9(Removable singularity).

Let $f$ be holomorphic in an open set $\Omega$, expect possibly at one point $z_0\in \Omega$. If we can define $f$ at $z_0$ in such a way that $f$ becomes holomorphic in all of $\Omega$, we say that $z_0$ is a removable singularity.

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Theorem 410.10(Riemann, Stein & Shakarchi (2003) 3.3.1).

Suppose that $f$ is holomorphic in an open set $\Omega$ except possibly at a point $z_0\in \Omega$. If $f$ is bounded on $\Omega \setminus \{z_0\}$, then $z_0$ is a removable singularity.

Proof.

We may consider a small disc $D$ centered at $z_0$ and whose closure is contained in $\Omega$. Let $C$ denote the boundary of $D$ with positive orientation. Define

$$g(z)=\frac{1}{2\pi i}{\int }_C\frac{f(\zeta )}{\zeta -z}d\zeta (z\in D)$$

Use Def 370.9:

$$\begin{array}{r}g(z)=\frac{1}{2\pi i}{\int }_0^{2\pi }\frac{f(z_0+ϵe^{i\theta })}{z_0+ϵe^{i\theta }-z}(ϵi{e}^{i\theta })d\theta .\end{array}$$

Thm 377.16 now shows that $g(z)$ is holomorphic on all of $D$.

We shall prove that if $z\in D$ and $z\ne z_0$, then under the assumptions of the theorem we have

$$f(z)=g(z)(z\in D\setminus \{z_0\})$$

Defining $f(z_0)=g(z_0)$ makes $f$ holomorphic at $z_0$, so $z_0$ is a removable singularity.

Fix $z\in D$ with $z\ne z_0$ and consider a keyhole contour with keyholes around $z_0$ and $z$. Letting the sides of the corridors get closer to each other and finally overlap, in the limit we get

$${\int }_C\frac{f(\zeta )}{\zeta -z}d\zeta +{\int }_{{\gamma }_{ϵ}}\frac{f(\zeta )}{\zeta -z}d\zeta +{\int }_{{\gamma }_{ϵ}^{\prime }}\frac{f(\zeta )}{\zeta -z}d\zeta =0,$$

where ${\gamma }_{ϵ}$ and ${\gamma }_{ϵ}^{\prime }$ are small circles of radius $ϵ$ with negative orientation centered at $z$ and $z_0$ respectively. Using Cauchy’s theorem, we have

$${\int }_{{\gamma }_{ϵ}}\frac{f(\zeta )}{\zeta -z}d\zeta =-2\pi if(z).$$

For the third integral, we use the assumption that $f$ is bounded and that since $ϵ$ is small, $\zeta$ stays away from $z$, and therefore

$$\left|{\int }_{{\gamma }_{ϵ}^{\prime }}\frac{f(\zeta )}{\zeta -z}d\zeta \right|⩽Cϵ.$$

Letting $ϵ\to 0$ proves our claim and concludes the proof.□

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Corollary 410.11(Stein & Shakarchi (2003) 3.3.2).

Suppose that $f$ has an isolated singularity at the point $z_0$. Then $z_0$ is a pole of $f$ iff $|f(z)|\to \infty$ as $z\to z_0$.

Proof.

If $z_0$ is a pole, then by definition $1/f$ has a zero at $z_0$, and therefore $|f(z)|\to \infty$ as $z\to z_0$. Conversely, suppose that this condition holds. Then, $1/f$ is bounded near $z_0$, and in fact $1/|f(z)|\to 0$ as $z\to z_0$. By Thm 10, $1/f$ has a removable singularity at $z_0$ and must vanish there. Thus, $z_0$ is a pole.□

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Definition 410.12(Essential singularity).

Isolated singularities that are not removable and are not poles are called essential singularities.

Thus, isolated singularities belong to one of these classes:

1. Removable singularities ($f$ bounded near $z_0$)
2. Pole singularities ($|f(z)|\to \infty$ as $z\to z_0$)
3. Essential singularities.

Definition 410.13(Singularities at infinity).

If a function is holomorphic of all large values of $z$, we can describe its behavior at infinity using the same tripartite distinction. If $f$ is holomorphic of all large values of $z$, we consider $F(z)=f(1/z)$, which is now holomorphic in a deleted neighborhood of the origin. We say that $f$ has a blank singularity at infinity if $F$ has a blank singularity at the origin. If $f$ has a removable singularity at $\infty$, we also say that $f$ is holomorphic at $\infty$.

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Contrary to the controlled behavior of a holomorphic function near a removable singularity or a pole, behavior near an essential singularity is typically more erratic.

Theorem 410.14(Casorati-Weierstrass, Stein & Shakarchi (2003) 3.3.3).

Suppose $f$ is holomorphic in the punctured disc $D_r(z_0)\setminus \{z_0\}$ and has an essential singularity at $z_0$. Then, the image of $D_r(z_0)\setminus \{z_0\}$ under $f$ is dense in $\mathbb{C}$.

Proof.

Assume that the range of $f$ is not dense, i.e, there exists $w\in \mathbb{C}$ and $\delta >0$ such that $|f(z)-w|>\delta$ for all $z\in D_r(z_0)\setminus \{z_0\}$. Then, we can define a new function on $D_r(z_0)\setminus \{z_0\}$ by

$$\begin{array}{r}g(z)=\frac{1}{f(z)-w},\end{array}$$

which is bounded by $1/\delta$ and is holomorphic on $D_r(z_0)\setminus \{z_0\}$. By Thm 10, $g$ has a removable singularity at $z_0$. If $g(z_0)\ne 0$, then $f(z)$ must have a removable singularity at $z_0$, a contradiction. If $g(z_0)=0$, then $f(z)-w$ has a pole at $z_0$, and $f$ has a pole at $z_0$ by Cor 11, also a contradiction.□

We now consider functions with only isolated singularities that are poles.

Definition 410.15(Meromorphic function).

A function $f$ on an open set $\Omega$ is meromorphic if there exists a sequence of points $Z=\{z_i{\}}_{i=0}^{\infty }$ that has no limit points in $\Omega$, and such that

1. the function $f$ is holomorphic in $\Omega \setminus Z$, and
2. $f$ has poles at each point in $Z$.

A meromorphic function in $\mathbb{C}$ that is either holomorphic at infinity or has a pole at infinity is said to be meromorphic in the extended complex plane.

Theorem 410.16(Stein & Shakarchi (2003) 3.3.4).

The meromorphic functions in the extended complex plane are the rational functions.

Proof.

Suppose $f$ is meromorphic in the extended complex plane. Then, $f(1/z)$ has a pole or a removable singularity at the origin. In either case, $f(1/z)$ is holomorphic in a neighborhood of the origin, which implies that all of the poles of $f$ (except infinity, of course) must lie in some closed disc of radius $r$ centered at the origin. Thus, by sequential compactness, $f$ can have only finitely many poles in the plane, say $z_1,\dots ,z_n$.

For each pole $z_k\in \mathbb{C}$, using Cor 5, there exists an open neighborhood $U_k$ in which we can write

$$f(z)=f_k(z)+g_k(z)$$

where $f_k(z)$ is the principal part of $f$ at $z_k$ and $g_k(z)$ is holomorphic in $U_k$. Note that $f_k$ is a polynomial in $1/(z-z_k)$. Similarly, we can write

$$f(1/z)={\tilde{f}}_{\infty }(z)+{\tilde{g}}_{\infty }(z),$$

where ${\tilde{g}}_{\infty }$ is holomorphic in a neighborhood of the origin and ${\tilde{f}}_{\infty }$ is the principal part of $f(1/z)$ at the origin (${\tilde{f}}_{\infty }=0$ if $f$ is holomorphic at infinity). Note that ${\tilde{f}}_{\infty }$ is a polynomial in $1/z$. Let $f_{\infty }(z)={\tilde{f}}_{\infty }(1/z)$. Thus, for all $z$ with $|z|>R>r$ for some $R$, we have

$$f(z)=f_{\infty }(z)+{\tilde{g}}_{\infty }(1/z).$$

Clearly, $|{\tilde{g}}_{\infty }(1/z)|$ is bounded in a neighborhood of infinity.

Define

$$\begin{array}{r}H:=f-f_{\infty }-\sum _{k=1}^n{f}_k.\end{array}$$

We contend that $H$ is entire and bounded. Indeed, near the poles $z_k$ we subtracted the principal part of $f$ so that the function $H$ has a removable singularity there. The summands of $H$ are clearly holomorphic on $\mathbb{C}\setminus \{z_1,\dots ,z_n\}$, so $H$ is entire. For all $z$ with large modulus, $f(z)-f_{\infty }(z)$ is bounded, and each $f_k$ is bounded outside $U_k$. Thus, $H$ is bounded, and by Liouville’s theorem we conclude that $H$ is constant. From the definition of $H$, we find that $f$ is the rational function $f_{\infty }+{\sum }_{k=1}^n{f}_k$. Since $\mathbb{C}$ is algebraically closed, every rational function is of this from, and is clearly meromorphic.□

Note that as a consequence, a rational function is determined up to a multiplicative constant by prescribing the locations and multiplicities of its zeroes and poles. (??)

The argument principle and applications

The argument principle provides a way to transform order data to residue data.

Theorem 410.17(Argument principle, Stein & Shakarchi (2003) 3.4.1).

Suppose $f$ is meromorphic in an open set containing a circle $C$ and its interior. If $f$ has no poles and never vanishes on $C$, then

$$\begin{array}{rl}\frac{1}{2\pi i}{\int }_C\frac{f^{\prime }(z)}{f(z)}dz& =\#(\text{zeroes of }f\text{ inside }C)-\#(\text{poles of }f\text{ inside }C),\end{array}$$

where the zeroes and poles are counted with their multiplicities.

The above theorem holds for toy contours.

Theorem 410.18(Rouchè, Stein & Shakarchi (2003) 3.4.3).

Suppose that $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior. If

$$|f(z)|>|g(z)|\forall z\in C,$$

then $f$ and $f+g$ have the same number of zeroes inside $C$.

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Theorem 410.19(Open mapping theorem, Stein & Shakarchi (2003) 3.4.4).

If $f$ is holomorphic and non-constant in a region $\Omega$, then $f$ is open.

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See Open Mapping Theorem (Complex Analysis) (2026)
Proof uses Thm 18.

Theorem 410.20(Maximum modulus principle, Stein & Shakarchi (2003) 3.4.5).

If $f$ is a non-constant holomorphic function in a region $\Omega$, then $f$ cannot attain a maximum in $\Omega$.

Proof uses Thm 19.

Corollary 410.21(Corollary).

Suppose that $\Omega$ is a region with compact closure $\overline{\Omega }$. If $f$ is holomorphic on $\Omega$ and continuous on $\overline{\Omega }$ then

$$\begin{array}{r}\underset{z\in \Omega }{\sup }|f(z)|⩽\underset{z\in \overline{\Omega }\setminus \Omega }{\sup }|f(z)|.\end{array}$$

In fact, since $f(z)$ is continuous on the compact set $\overline{\Omega }$, $|f(z)|$ attains its maximum in $\overline{\Omega }$; but this cannot be in $\Omega$ if $f$ is non-constant. If $f$ is constant, the conclusion is trivial.

Homotopies and simply connected domains

Proposition 410.22(Proposition).

If $f$ is holomorphic in $\Omega$, then

$$\begin{array}{r}{\int }_{{\gamma }_0}f(z)dz={\int }_{{\gamma }_1}f(z)dz\end{array}$$

whenever the two curves ${\gamma }_0$ and ${\gamma }_1$ are homotopic in $\Omega$.

Theorem 410.23(Theorem).

Any holomorphic function in a simply connected domain has a primitive.

Corollary 410.24(Corollary).

If $f$ is holomorphic in the simply connected region $\Omega ,$ then

$${\int }_{\gamma }f(z)dz=0$$

for any closed curve $\gamma$ in $\Omega$.

The complex logarithm

Theorem 410.25(Theorem).

Suppose that $\Omega$ is simply connected with $1\in \Omega$ and $0\notin \Omega$. Then in $\Omega$ there exists a holomorphic function $F(z)={\log }_{\Omega }(z)$, called a branch of the logarithm, so that

1. $F$ is holomorphic in $\Omega$;

2. $e^{F(z)}=z$ for all $z\in \Omega$;

3. $F(z)=\log r$ whenever $r$ is a real number near $1$.

Proof.

We construct $F$ as a primitive of the function $1/z$. Since $0\notin \Omega$, the function $f(z)=1/z$ is holomorphic in $\Omega$. We define

$${\log }_{\Omega }(z)=F(z)={\int }_{\gamma }f(w)dw,$$

where $\gamma$ is any curve in $\Omega$ connecting $1$ and $z$. Since $\Omega$ is simply connected, this definition does not depend on the path chosen. Arguing as before, we find that $F$ is holomorphic on $\Omega$ and $F^{\prime }(z)=1/z$.□

In the slit plane $\Omega =\mathbb{C}\setminus \{(-\infty ,0]\}$, we have the principal branch of the logarithm

$$\log z=\log r+i\theta$$

where $z=r{e}^{i\theta }$ with $|\theta |<\pi$. To prove this, integrate along the path that follows the $x$ axis till $x=r$ followed by an arc.

For the principal branch of the logarithm the following Taylor expression holds:

$$\log (1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\dots (|z|<1)$$

Indeed, the derivative of both sides equals $1/(1+z)$, so they must differ by a constant; they are both equal to $0$ at $z=0$.

Note that $\log (z_1{z}_2)\ne \log z_1+\log z_2$ in general.

Theorem 410.26(Theorem).

If $f$ is a nowhere vanishing holomorphic function in a simply connected region $\Omega$, then there exists a holomorphic function $g$ on $\Omega$ such that

$$f(z)=e^{g(z)}.$$

Proof.

Fix a point $z_0\in \Omega$ and define a function

$$g(z)={\int }_{\gamma }\frac{f^{\prime }(w)}{f(w)}dw+c_0,$$

where $\gamma$ is any path in $\Omega$ connecting $z_0$ to $z$ and $c_0$ is a complex number so that $e^{c_0}=f(z_0)$. Arguing as before, we find that $g$ is holomorphic with

$$g^{\prime }(z)=\frac{f^{\prime }(z)}{f(z)},$$

and a simple calculation gives

$$\frac{d}{dz}(f(z)e^{-g(z)})=0,$$

so that $f(z)e^{-g(z)}$ is a constant. Evaluating this expression at $z_0$, we find $f(z_0)e^{-c_0}=1$, so that $f(z)=e^{g(z)}$ for all $z\in \Omega$.□

Fourier series and harmonic functions

Suppose that $f$ is holomorphic in a disc $D_R(z_0)$, so that $f$ has a power series expansion

$$f(z)=\sum _{n=0}^{\infty }{a}_n(z-z_0{)}^n$$

that converges in that disc.

Theorem 410.27(Theorem).

The coefficients of the power series expansion of $f$ are given by

$$a_n=\frac{1}{2\pi r^n}{\int }_0^{2\pi }f(z_0+r{e}^{i\theta })e^{-in\theta }d\theta$$

for all $n⩾0$ and $0<r<R$. Moreover,

$$0=\frac{1}{2\pi r^n}{\int }_0^{2\pi }f(z_0+r{e}^{i\theta })e^{-in\theta }d\theta$$

whenever $n<0$.

Proof.

We know that $a_n=f^{(n)}(z_0)/n!$ from Thm 377.8; Just plug the parameterization $\zeta =z_0+r{e}^{i\theta }$ into Cor 377.6.□

Corollary 410.28(Corollary).

If $f$ is holomorphic in a disc $D_R(z_0)$, then

$$f(z_0)=\frac{1}{2\pi }{\int }_0^{2\pi }f(z_0+r{e}^{i\theta })d\theta 0<r<R.$$

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Harmonic functions

Definition 410.29(Harmonic function).

Let $u$ be a real-valued function defined on a disc $D$. We call $u$ harmonic if $u$ is twice continuously differentiable and $u_{xx}+u_{yy}=0$.

Proposition 410.30(Stein & Shakarchi (2003) Exr 2.12).

Let $u$ be harmonic on the unit disc $\mathbb{D}$. There exists a holomorphic function $f$ on the unit disc such that $\mathrm{Re}f=u$. The imaginary part of $f$ is uniquely determined up to an additive (real) constant.

Proof.

Motivated by Prp 370.5, define $g(z)=2\frac{\partial u}{\partial z}$. Show that $g$ satisfies the Cauchy-Riemann equations. Let a primitive of $g$ be $F$. If the real part of $F$ is $U$, we have

$$F^{\prime }=2\frac{\partial U}{\partial z}⟹\frac{\partial U}{\partial z}=\frac{\partial u}{\partial z}⟹U=u+c,$$

where $c\in \mathbb{R}$ is a constant. How define $f=F-c$, which works.

If $g$ is another holomorphic function satisfying the condition of the problem, then $\mathrm{Re}(f-g)=\mathrm{Re}(f)-\mathrm{Re}(g)=u-u=0$, which is constant, therefore $\text{Im}(f-g)$ is also constant.□

Taking the real parts of both sides in Cor 28, we have

Corollary 410.31(Corollary).

If $u$ is harmonic in a disc $D_R(z_0)$, then

$$u(z_0)=\frac{1}{2\pi }{\int }_0^{2\pi }u(z_0+r{e}^{i\theta })d\theta 0<r<R$$

Winding numbers

Definition 410.32(Definition).

For any closed path $\gamma$, we define its winding number with respect to a point $\alpha$ to be

$$W(\gamma ,\alpha )=\frac{1}{2\pi i}{\int }_{\gamma }\frac{1}{z-\alpha }dz,$$

provided the path does not pass through $\alpha$.

Lemma 410.33(Lemma).

If $\gamma$ is a closed path, then $W(\gamma ,\alpha )$ is an integer.

Warning

Homologous to 0 is not equivalent to homotopic to 0!

Theorem 410.34(Cauchy, Lang (2002) 4.2.2).

Let $\gamma$ be a closed chain in $U$, homologous to $0$ in $U$. Let $f$ be holomorphic in $U$. Then

$${\int }_{\gamma }f=0.$$

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Theorem 410.35(Lang (2002) 4.2.4).

Let $U$ be an open set and $\gamma$ a closed chain in $U$ such that $\gamma$ is homologous to $0$ in $U$. Let $z_1,\dots ,z_n$ be a finite number of distinct points of $U$. Let ${\gamma }_i$ ($i=1,\dots ,n$) be the boundary of a closed disc ${\overline{D}}_i$ contained in $U$, containing $z_i$, and oriented counterclockwise. We assume that ${\overline{D}}_i$ does not intersect ${\overline{D}}_j$ if $i\ne j$. Let $m_i=W(\gamma ,z_i)$. Let $U^{\ast }$ be the set obtained by deleting $z_1,\dots ,z_n$ from $U$. Then $\gamma$ is homologous to $\sum m_i{\gamma }_i$ in $U^{\ast }$.

It follows from Thm 34 that if $f$ is holomorphic on $U^{\ast }$, then

$${\int }_{\gamma }f=\sum _{i=1}^n{m}_i{\int }_{{\gamma }_i}f.$$

We can now state a generalization of Thm 377.5.

Theorem 410.36(Cauchy's Formula, Lang (2002) 4.2.5).

Let $\gamma$ be a closed chain in $U$, homologous to $0$ in $U$. Let $f$ be holomorphic on $U$, let $z$ be in $U$ and not on $\gamma$. Then

$$\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(\zeta )}{\zeta -z}d\zeta =W(\gamma ,z)f(z).$$

Laurent expansions

[!Theorem] Lang (2002) 5.2.1

I want to calculate the integral of $\frac{1}{z(z-1)}$ over a centered at the origin of radius 2. I first split the fraction: $\frac{1}{z-1}-\frac{1}{z}$. I can then use Cauchy’s theorem to integrate over the chain which consists of two tiny circles centered at $0$ and $1$. The terms alternatively vanish, and the one that remains contributes its residue. Since the residues are $1$ and $-1$, the integral must be $0$.

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References

Lang, S. (2002). Algebra (Vol. 211). Springer New York. https://doi.org/10.1007/978-1-4613-0041-0

Open Mapping Theorem (Complex Analysis). (2026). Wikipedia. https://en.wikipedia.org/w/index.php?title=Open_mapping_theorem_(complex_analysis)&oldid=1336386408

Stein, E. M., & Shakarchi, R. (2003). Complex Analysis. Princeton University Press.