Category Archives: Complex Analysis

Automorphism of annulus

Suppose \mathbb{R}_\mu=\{z\in \mathbb{C}:1<z<\mu\} is the annulus for \mu>1. Then

\bf{Thm:}\mathbb{R}_\mu is not biholomorphic to \mathbb{R}_{\mu'} whenever \mu\neq \mu'.

The proof goes like following: Suppose there exists a biholomorphic map between them, then this map can be extended to the boundary by the so called Kellog theorem, then one can reflect the annulus the inner part and outside part. Near the origin, one can use the removable singularity theorem to prove the extended map is holomophic. Thus we can get a biholomorphic map between \mathbb{C} and \mathbb{C} which must have the form az+b. First b=0, because the map keeps the origin. Second a must have the unit length to keep the inner boundary to be |z|=1. Thus the map can only be rotation around origin.

This actually tells us the automorphism of an annulus can only be the rotation. Let us verify this fact by another perspective.


As we all know, \mathbb{R}_\mu is biholomorphic to \mathbb{H}/\Gamma_\lambda, where \mathbb{H} is the upper half plane and \Gamma_\lambda(z)=\lambda z. Here \ln \lambda\ln \mu=2\pi^2. \Gamma_\lambda acts on the \mathbb{H} freely discontinuously.

Any automorphism of \mathbb{H}/\Gamma_\lambda can be lift to an automphism of \mathbb{H} which commutes with the action \Gamma_\lambda. In the matrix language, these automphisms are the matrix in PSL(2,\mathbb{R}) commutes with \left(\begin{matrix}\sqrt{\lambda}&0\\0&\sqrt{\lambda^{-1}}\end{matrix}\right). Some elementary calculation shows that these matrix have the form \left(\begin{matrix}a&0\\0&a^{-1}\end{matrix}\right) where a\in \mathbb{R}. Any one of these  is corresponding to a rotation of the annulus.



Polynomial on each variable

Problem: Suppose {f\in \mathcal{O}(\mathbb{C}^2)}. If {f(z_1,z_2)} is a polynomial of {z_1} for every fixed {z_2} and a polynomial of {z_2} for every fixed {z_1}. Then {f} must be a polynomial.

Proof: Since {f} is holomorphic on the {\mathbb{C}^2}, we have the power series expansion

\displaystyle f(z_1,z_2)=\sum_{i,j\geq 0}a_{ij}z_1^iz_2^j=\sum_{i\geq 0}a_i(z_1)z_2^i

where {a_i(z_1)} is a holomorphic function on {\mathbb{C}} for any {i\geq 0}. Suppose there exists a sequence {i_1,i_2,\cdots} with {\lim_{k\rightarrow \infty}i_k=\infty} satisfying {a_{i_k}(z_1)\neq 0}. Since the set of roots of {a_{i_k}} are countable in {\mathbb{C}}, then all the union of these sets are also countable. Therefore, {\exists\, w\in\mathbb{C}} such that {a_{i_k}(w)\neq 0} for all {k\geq 0}. Then {f(w,z_2)} can never be a polynomial. So

\displaystyle f(z_1,z_2)=\sum_{i=0}^Ma_i(z_1)z_2^i=\sum_{i\geq 0}b_i(z_2)z_1^i

where every {b_i} is polynomial of {z_2}. Repeating the above proof we get this is also a finite sum, which means {f} is a polynomial. \Box

Remark: Exercise from Shabat, 1.10. Idea comes from Hanlong Fang.

Some uniqueness sets of holomorphic function

Suppose {f\in \mathcal{O}(\mathbb{C}^2)}, if {f} is zero on the following set
(a) A real hyperplane in {\mathbb{C}^2}
(b) The real two dimensional plane {\{z_1=\bar{z}_2\}}
(c) the arc {\{z_1=\bar{z}_2,y_1=x_1\sin(1/x_1)\}}
Show that {f} must be zero on {\mathbb{C}^2}.

Proof: (a) Suppose the hyperplane has normal vector {(1,0)\in \mathbb{C}^2} and passing through origin, then {(iy,z_2)} will be in this hyperplane for any {y\in \mathbb{R}}.

Fix {z_2}, {f(\cdot,z_2)} is a holomorphic function in {\mathbb{C}}. Because zero point of holomorphic function is isolated unless a constant, {f(iy,z_2)=0\quad\forall\,y\in \mathbb{R}} will imply {f(z_1,z_2)\equiv 0}.

(b) Let {w_1=\frac{z_1+z_2}{2}}, {w_2=\frac{z_1-z_2}{2i}}, then

\displaystyle \{z_1=\bar{z}_2\}=\{\text{Im}\,w_1=\text{Im}\,w_2=0\}

{g(w_1,w_2)=f(z_1,z_2)} is also a holomorphic function in {w_1,w_2}. As we did in part (a), fix {w_1=x\in \mathbb{R}}, {g(x,y)=0} imply {g(x,w_2)=0} for any {w_2\in\mathbb{C}}. Since {x} is arbitrary, then {g(w_1,w_2)\equiv 0}.

(c) Use the transformation of part (b), {\{z_1=\bar{z}_2, y_1=x_1\sin \frac{1}{x_1}\}} is equivalent

\displaystyle \gamma=\{\text{Im}\,w_1=\text{Im}\,w_2=0,w_2=w_1\sin\frac{1}{w_1}\}

Consider {g(w_1,w_2)}, then {g(\cdot,0)=0} has a sequence of roots {w_1} converging to 0, then {g(w_1,0)=0} for any {w_1\in \mathbb{C}}. From the Taylor expansion, there exists {g_1\in \mathcal{O}(\mathbb{C}^2)} such that

\displaystyle g(w_1,w_2)=w_2g_1(w_1,w_2)

Since {g(w_1,w_2)} is zero on {\gamma}, then {g_1(w_1,w_2)} is zero on {\gamma} when {w_2\neq 0}. By continuity, {g_1(w_1,w_2)} is zero on {\gamma}.

Then we can repeat the above procedure, {\exists\, g_2} such that

\displaystyle g_1(w_1,w_2)=w_2g_2(w_1,w_2),\quad g=w_2^2g_2

Then we must have {g\equiv 0}, otherwise we can continue this process infinitely. \Box

Remark: The idea of part (b) is due to Lun Zhang.


Complex lines in C^2

Problem: (a) Suppose {l_1} and {l_2} are two complex lines in {\mathbb{C}^2} which are not orthogonal. Prove that the projection of circles {\Gamma\subset l_1} to {l_2} are also circles.

(b) Suppose {l_1} is a 2 dimensional real plane in {\mathbb{C}^2} and {l_2} is a complex line. If for any circle {\Gamma\subset l_1}, its projection to {l_2} is a circle on {l_2}, then {l_1} must be a complex or anticomplex line. Proof: (a) Under some translation and rotation which preserves the projection and circles, we can assume

\displaystyle l_1=\{(z,wz)|z\in \mathbb{C}\}

\displaystyle l_2=\{(\xi,0)|\xi\in \mathbb{C}\}

where {w} is a fixed constant in {\mathbb{C}}. Here we have used the property that {l_2} is not orthogonal to {l_1}. Suppose {\Gamma} is circle in {l_1} with center {(z_0, wz_0)} with radius {r}, then points on {\Gamma} satisfy

\displaystyle r=|(z-z_0,w(z-z_0))|=|z-z_0|\sqrt{1+|w|^2}

which means {|z-z_0|=const}. This exactly means the projection on {l_2} is a circle.

(b) Any 2 dimensional real space through origin can be respresented by

\displaystyle l_1=\{\alpha\zeta+\beta\bar{\zeta}|\zeta\in \mathbb{C}\}

where {\alpha=(\alpha_1,\alpha_2)\in\mathbb{C}^2,\beta=(\beta_1,\beta_2)\in\mathbb{C}^2}. WLOG, assume {l_2} is the same as (a).

If {|\alpha_1|\neq |\beta_1|}, then let {z=\alpha_1\zeta+\beta_1\bar{\zeta}},

\displaystyle l_1=\left\{\left(z,\frac{(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}{|\alpha_1|^2-|\beta_1|^2}\right)\right\}

By projecting to {l_2}, we get {|z|=const} must imply {(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}} has constant norm also. One can verify that this is only true when {\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2=0} or {\alpha_1\beta_2-\alpha_2\beta_1=0}, which means {l_1} must be an anti-complex line or complex line.

If {|\alpha_2|\neq |\beta_2|}, we can porject circles on {l_1} to {l_3=\{(0,\zeta)\}}, which are also circles by assumption. So we can apply all the above proof.

Next suppose {|\alpha_1|=|\beta_1|\neq 0} and {|\alpha_2|= |\beta_2|\neq 0}, first note that

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}\equiv const

In fact, for any {\zeta}, there exists {c_1>0} such that

\displaystyle |\alpha_1\zeta+\beta_1\bar{\zeta}|+|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1

then {\zeta/c_1} will be in a circle in {l_1} whose center is origin and radius 1. So {|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1c_0} for some constant {c_0}. Then

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{c_1c_0}{c_1-c_1-c_0}=const

Choosing {\zeta} in particular, this constant must be {|\alpha_1|/|\alpha_2|=|\beta_1|/|\beta_2|}. From

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{|\alpha_1|}{|\alpha_2|}

Squaring it and write it as equality, we get

\displaystyle [|\alpha_2|^2\alpha_1\bar{\beta}_1-|\alpha_1|^2\alpha_2\bar{\beta}_2]\zeta^2=[|\alpha_1|^2\bar{\alpha_2}\beta_2-|\alpha_2|^2\bar{\alpha_1}\beta_1]\bar{\zeta}^2


\displaystyle |\alpha_2|^2\alpha_1\bar{\beta}_1=|\alpha_1|^2\alpha_2\bar{\beta}_2

\displaystyle |\alpha_1|^2\bar{\alpha_2}\beta_2=|\alpha_2|^2\bar{\alpha_1}\beta_1

This two imply {\alpha_1/\beta_1=\alpha_2/\beta_2=\eta^{-1}}. Hence

\displaystyle l_1=\{(\alpha_1,\alpha_2)(\zeta+\eta\bar{\zeta})|\zeta\in\mathbb{C}\}

However, {|\zeta|=|\eta\bar{\eta}|} implies that

\displaystyle \text{Arg}\,(\zeta+\eta\bar{\zeta})=\frac{1}{2}\text{Arg}\,\eta

then {l_1} is in fact only one dimesional in this case. Contradiction. \Box

Remark: Shabat, Introduction to complex analysis Part II. Functions of several variables.

Hadamard three lines lemma

\mathbf{Problem:} Let stripe S=\{z\in \mathbb{C}|0<\Re(z)<1\}, f is analytic in S and continuous to the boundary of S. Suppose |f(z)|\leq e^{C|z|^{2-\delta}} for some C>0 and \delta>0. Define \displaystyle M(x)=\sup\limits_{y}|f(x+iy)|. Then M(t)\leq M(0)^{1-t}M(1)^{t} for 0\leq t\leq 1, which means logM(t) is a convex function.

\mathbf{Proof:} First let us prove if M(0)\leq 1 and M(1)\leq 1, then |f|\leq 1.

Fix \epsilon>0, define F(z)=e^{\epsilon z^2}f(z) on S, then |F(z)|\to 0 as z\to \infty. By the maximum principle, \sup_S|F(z)| must occur on the edge of S.

|F(z)|\leq e^{\epsilon z^2} for z\in S

Let \epsilon \to 0, we get |f(z)|\leq 1 for z\in S.

For the general case, define F(z)=f(z)M(0)^{z-1}M(1)^{-z}, then |F(z)|\leq 1 on \partial S. And F has growth rate no more than some e^{C|z|^{2-\delta}}, so apply the above result, we know that \sup_S |F(z)|\leq 1, which means \sup_S |f(z)|\leq M(0)^{1-z}M(1)^{z}.

Let z=t\in [0,1], we get the conclusion.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} f can not grow too fast, otherwise this theorem fails.

Consider \displaystyle f(z)=e^{e^{\pi(z-\frac 12)i}} on S. Easy to verify that |f(z)|\leq 1 on \partial S. But on \Re(z)=\frac 12, f grows extremely fast.

Torus and Moduli space

Let w_1, w_2\in \mathbb{C}, define group G\subset \text{Aut}(\mathbb{C}) generated by two operations

z\longrightarrow z+w_1

z\longrightarrow z+w_2

Assume w_1, w_2 are real linear independent, then \mathbb{C}/G is called a torus. In general we would orient the torus by assuming  \displaystyle \text{Im}\frac{w_2}{w_1}>0.

\mathbf{Proposition:}  \mathbb{C}/G_1 and \mathbb{C}/G_2 are conformally equivalent if and only if G_1 and G_2 are conjugate in \text{Aut}(\mathbb{C}).

\mathbf{Proof:} If G_1 and G_2 are conjugate then it is easy to prove the conclusion. 

 Suppose \phi: \mathbb{C}/G_1\cong \mathbb{C}/G_2 and \pi_i:\mathbb{C}\to \mathbb{C}/G_i. Since the universal cover of \mathbb{C}/G_i is \mathbb{C}, by the lifting lemmam, there exists a unique \tilde{f}: \mathbb{C}\to \mathbb{C} such that the following diagram commutes

From f^{-1}, \exists ! \psi. Then \tilde{f}\psi=\psi\tilde{f}=id, which means \tilde{f}\in \text{Aut}(\mathbb{C}).

Suppose g_1\in G_1, then from \pi_2\circ \tilde{f}=f\circ \pi_1,  \pi_2\tilde{f}g_1(z)=f \pi_1(z)=\pi_2\tilde{f}(z). So there exists \tau_z\in G_2 such that \tilde{f}g_1(z)=\tau_z(\tilde{f}(z)). Actually, the choice of \tau_z does not depend on z. This is because the group G_1 and G_2 act discontinouly on \mathbb{C}, then there is a neighborhood of z such that \tau_z remain the same.

So fix z_0\in\mathbb{C}, \displaystyle S=\{z|\tau_z=\tau_{z_0}\} is an open set. S is also closed, since G_1,G_2 are compatible with the topology of \mathbb{C} and f,\tilde{f} are continuous. So S=\mathbb{C}, which means for a fixed g_1, \exists\, \tau\in G_2 such that \tilde{f}\circ g_1=\tau\circ \tilde{f}. Thus \tilde{f}G_1\tilde{f}^{-1}\subset G_2. For the same sake, we have \tilde{f}^{-1}G_2\tilde{f}\subset G_1. G_1 and G_2 are conjugate in \text{Aut}(\mathbb{C}). \text{Q.E.D. }\hfill \square

Consider \mathbb{C}/G_1\cong \mathbb{C}/G_2. Suppose G_1 is generated by the mobius transformation z\to z+w_1 and z\to z+w_2, G_2 is generated by z\to z+\tau_1 and z\to z+\tau_2. Let us denote this four mobius transformations as w_1, w_2,\tau_1,\tau_2 and \displaystyle w=\frac{w_1}{w_2}, \displaystyle \tau=\frac{\tau_1}{\tau_2}.

Since \text{Aut}(\mathbb{C})=\{az+b|a\neq 0\}, \forall\, f=az+b, fG_1f^{-1}=\langle aw_1,aw_2\rangle. Then \exists\,k,l,m,n\in \mathbb{Z} such that



Then \displaystyle w=\frac{k\tau+l}{m\tau+n}.  Since w and \tau are symmetric, \begin{pmatrix} k &l\\m & n \end{pmatrix} must have inverse in \text{Mat}_{2}(\mathbb{Z}). Thus \det \begin{pmatrix} k &l\\m & n \end{pmatrix}=\pm 1. Since we require \text{Im}\, w and \text{Im}\, \tau be positive, then \det \begin{pmatrix} k &l\\m & n \end{pmatrix}=1.

\mathbf{Thm:} \mathbb{C}/G_1 and \mathbb{C}/G_2 are conformally equivalent if and only if \displaystyle \frac{w_1}{w_2}\sim \frac{\tau_1}{\tau_2} in PSL(2,\mathbb{Z}). The module space of torus is PSL(2,\mathbb{Z}).

Uniformization of Riemann Surface and their automorphisms

\mathbf{Thm1:} Every simply connected Riemann surface must comformally equivalent to one of the followings:

  • The Riemann sphere, \hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}
  • The complex plane, \mathbb{C}
  • The unit disk. \mathbb{D}=\{z\in \mathbb{C}||z|<1\}

\mathbf{Thm2:} Every Riemann surface M is conformally equivalent to \Sigma/G where \Sigma is one of standard types, \hat{\mathbb{C}}, \mathbb{C}, \mathbb{D}. G is a subgroup of \text{Aut}(\Sigma) that acts freely discontinuously on \Sigma. Furthermore G\cong \pi_1(M).

\mathbf{Thm3:} Two Riemann surfaces are conformally equivalent if and only if they have the same \Sigma and their G are conjugate in \text{Aut}(\Sigma).

Basic facts:

\text{Aut}(\hat{\mathbb{C}})= Mobius  transformations=PSL(2, \mathbb{C})= SL(2,\mathbb{C})/{\pm I}

\text{Aut}({\mathbb{C}})=\{az+b|a, b\in \mathbb{C}, a\neq 0\}=\text{Aff}(1,\mathbb{C})

\text{Aut}({\mathbb{D}})\cong PSL(2,\mathbb{R})

Some generalizations of theorems of holomorphic function.

\mathbf{Thm:} If f\not\equiv constant is analytic in a domain U, then f(U) is also a domain. If f only has poles in U, this conclusion still holds.

\mathbf{Thm:} f(z) holomorphic at z_0\in \overline{\mathbb{C}}, where it has a zero point of order k\geq 1, then there is a neighborhood of \mathcal{N}(z_0) and a neighborhood of \mathcal{N}(w_0) such that every point in \mathcal{N}(w_0) except w_0 itself has precisely k distinct inverse images in \mathcal{N}(z_0). If f has pole of order k at z_0, conclusion still holds.

\mathbf{Proof:} Suppose f has pole \phi(z)=\frac{1}{f(z)}, \eta=\frac 1 w. Observe that w=\frac {1}{\eta} carries |\eta|<\delta into the disc |w|>\frac 1\delta. If z_0=\infty, consider both transformations \xi=\frac 1z, \eta=\frac 1 w. \square

\mathbf{Corollary:} If w=f(z) is univalent(injective) on a domain G, then all the zeros and poles of f(z) in G  are simple.

The First Lemma in Riemann Mapping Theorem

\mathbf{Problem:} Suppose U is a simply connected domain in \mathbb{C}, U\neq \mathbb{C}. Then there exists a holomorphic function f:U\to \mathbb{D}, the unit disc, and f is injective.
\mathbf{Proof:} Suppose z_0\not\in U, then g(z)=z-z_0 never achieves zero on U. Since U is simply connected, there exists a holomorphic function h such that h^2(z)=g(z), \forall x\in U.
Choose a\in U, then h(a)=b\neq 0. From open map theorem, \exists\, r>0 such that B_r(b)\subset h(U). Then B_r(-b)\cap h(U)=\emptyset, otherwise \exists\, a_1\in B_r(b), a_2\in B_r(-b) such that h(a_1)=-h(a_2). Since g=h^2 is injective, one must have a_1=a_2, so B_r(b)\cap B_r(-b)\neq \emptyset, which means 0\in B_r(b)\cap B_r(-b)\in h(U). Contradiction.
So one can check

\displaystyle f(z)=\frac{r}{h(z)+b}

satisfies all the conditions.


  • Pay attention to the domain U=\mathbb{C}\backslash (-\infty, 0], which is simply connected. Some other proof may have flaws when dealing with this stuition.
  • Riemann Mapping Theorem was firstly stated in his PhD thesis. Riemann’s flawed proof depended on the Dirichlet principle (whose name was created by Riemann himself).  Karl Weierstrass found that this principle was not universally valid. Later, David Hilbert was able to prove that, to a large extent, the Dirichlet principle is valid under the hypothesis that Riemann was working with However, in order to be valid, the Dirichlet principle needs certain hypotheses concerning the boundary of U which are not valid for simply connected domains in general. See Riemann Mapping Theorem.

Pringsheim Theorem

\mathbf{Thm(Pringsheim):} Suppose f(z)=\sum\limits_{n=0}^{\infty}a_nz^n has convergence radius 1. If all a_n\geq 0, then z=1 is a singular point of f.

\mathbf{Proof:} Expand f as Taylor series at z=\frac 12

\displaystyle f(z)=\sum \limits_{k=0}^{\infty}(z-\frac 12)^k\sum\limits_{n=k}^{\infty}C_n^ka_n(\frac 12)^{n-k}

If f has holomorphic extension on z=1, then the above series must converge for some z=1+\epsilon, \epsilon>0, small enough,

\displaystyle f(1+\epsilon)=\sum \limits_{k=0}^{\infty}(\frac 12+\epsilon)^k\sum\limits_{n=k}^{\infty}C_n^ka_n(\frac 12)^{n-k}

Since the double series has positive terms, we can interchange the order of two summations without loss of convergence. Hence the series,

\displaystyle\sum \limits_{n=0}^{\infty}a_n\sum\limits_{k=0}^{n}C_n^ka_n(\frac 12+\epsilon)^k(\frac 12)^{n-k}=\sum \limits_{n=0}^{\infty}a_n(1+\epsilon)^n

must converge. This contradicts the fact that f has convergence radius 1. \square

  • This theorem is posted by Alfred Pringsheim. But his proof is not right. For correct proof, refer to Analytic function Volume 1 by Einar Hille.
  • This theorem concludes there is at least one singular point fo the power series on its circle of convergence.
  • Consider f_p(z)=\sum\limits_{n=0}^{\infty}\frac{1}{n^p}z^n, p is real, which has convergence radius 1. This theorem applies and shows that z=1 is a singular point of f_p(z). actually, it is the only singularity on |z|=1. In this case, f_p(z)\to \infty, as z\to 1 when p\leq 1 but not for p>1. f^{(k)}(z)\to \infty as z\to 1 when p\leq k+1 but not for p>k+1.