# Some solutions to Rudin's complex analysis book

The following notebook contains some solutions to the complex analysis part of the Big Rudin book that I studied at POSTECH. This post is also a chance for me to test the different between MathJax and KaTeX in Nikola, to see which one has better render. It turns out that KaTeX is much faster than MathJax. As a note, to make KaTeX work with inline mode from Jupyter notebook, we have to change the default auto-render code in the theme (as suggested in this file, the `renderMathInElement`

part).

- Chapter 10: Elementary Properties of Holomorphic Functions
- Chapter 11: Harmonic Functions
- Chapter 12: The Maximum Modulus Principle
- Chapter 13: Approximation by Rational Functions
- Chapter 14: Conformal Mapping

### Chapter 10 - Elementary Properties of Holomorphic Functions¶

**1.** The following fact was tacitly used in this chapter: If $A$ and $B$ are disjoint subsets of the plane, if $A$ is compact, and if $B$ is closed, then there exists a $\delta > 0$ such that $|\alpha - \beta| \geq \delta$ for all $\alpha \in A$ and $\beta \in B$. Prove this, with an arbitrary metric space in place of the plane.

**Proof.** Let $A$ be a compact set and $B$ be a closed set in a metric space such that $A\cap B = \varnothing$. Let $\delta = \inf_{\alpha,\beta}d(\alpha,\beta)$ where the infimum is taken with all $\alpha \in A$ and $\beta\in B$, and let $\{a_n\}$, $\{b_n\}$ be two sequences in $A$ and $B$ correspondingly such that $\lim_{n\to\infty}d(a_n,b_n) = \delta$. Suppose that $\delta = 0$. Because $A$ is compact, there exists $c\in A$ and a subsequence $\{a_{n_k}\}$ of $\{a_n\}$ such that $\lim_{k\to\infty}d(a_{n_k},c) = 0$. Hence $\lim_{k\to\infty}d(b_{n_k},c) = 0$ because $\lim_{k\to\infty}d(a_{n_k},b_{n_k}) = 0$. Hence $c \in \bar{B}=B$, which is a contradiction with the hypothesis that $A\cap B = \varnothing$. So $\delta > 0$. We get the conclusion. $\Box$

**2.** Suppose that $f$ is an entire function, and that in every power series
$$
f(z) = \sum_{n=0}^{\infty} c_n(z-a)^n
$$
at least one coefficient is $0$. Prove that $f$ is a polynomial.

*Hint.* $n!c_n = f^{(n)}(a)$.

**Proof.** Let $a\in \mathbb{C}$. Consider the power series of $f$ at $a$, we have
$$
f(z) = \sum\limits_{n=0}^{\infty}c_n(z-a)^n.
$$

By assumption, $c_n = 0$ for some $n\leq 0$. By $n!c_n = f^{(n)}(a)$, we get $f^{(n)}(a) = 0$ for some $n\leq 0$. Put $K = \{z:f^{(n)}(z) = 0\text{ for some }n\leq 0\}$, we have $a \in K$. This is true for every $a\in \mathbb{C}$. So $K = \mathbb{C}$. Moreover, we have $$ K = \bigcup\limits_{n=0}^{\infty}\{z:f^{(n)}(z) = 0\}. $$

If, for each $n$, we do not have $f^{(n)}(z) = 0$ for every $z\in \mathbb{C}$, then $\{z:f^{(n)}(z) = 0\}$ is a countable set for all $n$. Hence $K$ is a countable set, which is a contradiction to $K = \mathbb{C}$. So $f^{(N)} = 0$ for some $N\geq 0$. Consider the power series of $f$ at $0$, we have $$ f(z) = \sum\limits_{n=0}^{\infty}b_n z^n. $$

By $f^{(n)}(0)=0$ for all $n\geq N$, we have $b_n = f^{(n)}(0)/n! = 0$ for all $n\geq N$. We get the conclusion. $\Box$

**3.** Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for every $z$. What conclusion can you draw?

**Proof.** $f = Cg$, where $|C| \leq 1$.

If $g$ is not vanishing in $\mathbb{C}$, then $Z(g)$ has no limit point. Put $h = f/g$, then $h \in \mathbb{C}\backslash Z(g)$ and $h$ is a meromorphic function in $\mathbb{C}$. Let $a\in Z(g)$. By assumption, $|h|\leq 1$ in a deleted disc $D'(a,r)$ for some $r >0$ such that $g\neq 0$ in $D'(a,r)$. So $h$ has a removable singularity at $a$. This is true for all $a\in Z(g)$, hence $h\in H(\mathbb{C})$ and $|h|\leq 1$ in $\mathbb{C}$. By Liouville's theorem, $h$ is a constant $C$ in $\mathbb{C}$. So we get $f = Cg$ and it is clear that $|C| \leq 1$. For the case $g =0$ in $\mathbb{C}$, we get $f = 0$ in $\mathbb{C}$, hence $f= Cg$ with $C = 0$. Reversely, if $f=Cg$ and $|C|\leq 1$, then $|f|\leq |C||g| \leq |g|$. $\Box$

**7.** If $f\in H(\Omega)$, the Cauchy formula for the derivatives of $f$,
$$
f^{(n)}(z) = \frac{n!}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{(\zeta - z)^{n+1}}\,d\zeta\qquad (n=1,2,3,\ldots)
$$
is valid under certain conditions on $z$ and $\Gamma$. State these, and prove the formula.

**Proof.** $\Gamma$ is a cycle in $\Omega$, $\mathrm{Ind}_{\Gamma}(a)=0$ for all $a\notin \Omega$, $z\in \Omega\backslash \Gamma$, and $\mathrm{Ind}_{\Gamma}(z) = 1$.

By Cauchy's formula, with the above conditions, we have $$ f(w) = \frac{1}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{\zeta-w}\,d\zeta\qquad (w\in\Omega\backslash \Gamma). $$

Let $r > 0$ such that $D(z;r)\subset \Omega\backslash\Gamma$. For $w \in D'(z;r/2)$, we have $$ \frac{f(w)-f(z)}{w-z} = \frac{1}{2\pi i(w-z)}\int_{\Gamma}f(\zeta)\left(\frac{1}{\zeta-w} - \frac{1}{\zeta - z}\right)\,d\zeta. $$

Let $w\to z$, we get $$ \lim_{w\to z}\frac{1}{w-z}\left(\frac{1}{\zeta-w} - \frac{1}{\zeta - z}\right) = \left(\frac{1}{\zeta - z}\right)' = \frac{1}{(\zeta - z)^2}. $$

Moreover, for all $w \in D'(z;r/2)$, we have $$ \left|\frac{f(\zeta)}{w-z}\left(\frac{1}{\zeta-w} - \frac{1}{\zeta - z}\right)\right| = \left|\frac{f(\zeta)}{(\zeta-w)(\zeta-z)}\right|\leq \frac{4}{r^2} \|f \|_{\Gamma}\qquad(\zeta \in \Gamma). $$

An easy argument by applying the dominated convergence theorem shows that $$ f'(z) = \lim_{w\to z}\frac{f(w)-f(z)}{w-z} =\frac{1}{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta. $$

We have proved the required formula for the case $n=1$. Note that $$ \lim_{w\to z}\frac{1}{w-z}\left(\frac{1}{(\zeta-w)^n} - \frac{1}{(\zeta - z)^n}\right) = \left(\frac{1}{(\zeta - z)^n}\right)' = \frac{1}{(\zeta - z)^{n+1}} $$ and for all $w \in D'(z;r/2)$, we have $$ \begin{aligned} \left|\frac{f(\zeta)}{w-z}\left(\frac{1}{(\zeta-w)^n} - \frac{1}{(\zeta - z)^n}\right)\right| &= \left|\frac{f(\zeta)\sum_{i=0}^{n-1}(\zeta-w)^i(\zeta-z)^{n-i-1}}{(\zeta-w)^n(\zeta -z)^n}\right| \\ &=\left|f(\zeta)\sum\limits_{i=0}^{n-1}\frac{1}{(\zeta-w)^{n-i}(\zeta-z)^{i}}\right|\\ &\leq \frac{2^n n}{r^n}\|f\|_{\Gamma}\qquad (\zeta\in \Gamma). \end{aligned} $$

By an induction procedure, the above notes, and a similar argument as the case $n=1$, we can prove the required formula for arbitrary $n \geq 1$. $\Box$

**19.** Suppose $f\in H(U)$, $g\in H(U)$, and neither $f$ nor $g$ has a zero in $U$. If
$$
\frac{f'}{f}\left(\frac{1}{n}\right) = \frac{g'}{g}\left(\frac{1}{n}\right)\qquad (n=2,3,\ldots),
$$
find another simple relation between $f$ and $g$.

**Proof.** $f=Cg$, where $C\neq 0$.

Put $h = f/g$. It is clear that $h\in H(U)$. Moreover, by $$ h'(z) = \frac{f'(z)g(z) - f(z)g'(z)}{g^2(z)}, $$ we get $h'(\frac{1}{n})=0$ for $n=2,3,\ldots$ Because $\frac{1}{n}\to 0$ as $n\to \infty$, we get $h'(z) = 0$ for every $z\in U$. So $h(z) = h(0) + \int_{[0,z]}h'(w)\,dw = h(0)$ for all $z \in \mathbb{C}$. Put $C = h(0)=f(0)/g(0)\neq 0$, we get $f = Cg$. Reversely, if $f = Cg$ and $C\neq 0$, we get $f'(z)/f(z) = g'(z)/g(z)$ for all $z\in U$, in particularly, for $\frac{1}{n}$, $n=2,3,\ldots$ $\Box$

**21.** Suppose $f\in H(\Omega)$, $\Omega$ contains the closed unit disc, and $|f(z)| < 1$ if $|z|=1$. How many fixed points must $f$ have in the disc? That is, how many solutions does the equation $f(z)=z$ have there?

**Proof.** 1 solution.

Let $g$ and $h$ be two functions in $\Omega$ defined by $g(z) = f(z) - z$ and $h(z) = -z$ for every $z\in \Omega$. By assumption, $ |g(z) - h(z)| < |h(z)| $ for all $z$ in the circle $\{z:|z|=1\}$. By Rouché's theorem, we get the number of solutions of $g$ in $D(0;1)$ equals to the number of solutions of $h$ in $D(0;1)$, which contains just one solution $z=0$. We get the conclusion. $\Box$

### Chapter 11 - Harmonic Functions¶

**1.** Suppose $u$ and $v$ are real harmonic functions in a plane region $\Omega$. Under what conditions is $uv$ harmonic? (Note that the answer depends strongly on the fact that the question is one about real functions.) Show that $u^2$ cannot be harmonic in $\Omega$, unless $u$ is constant. For which $f\in H(\Omega)$ is $|f|^2$ harmonic?

**Proof.** $u$ is constant or $v$ is constant or there is some real constant $C\neq 0$ such that $u_x = Cv_y$, $u_y = -Cv_x$ (in other words, $u+iCv$ is holomorphic). $f$ is constant.

Note that every harmonic function has continuous partial derivatives of all orders. Suppose $uv$ is harmonic, which means $(uv)_{xx}+(uv)_{yy}=0$ in $\Omega$, hence $u_xv_x + u_yv_y = 0$ in $\Omega$ (because $u_{xx}+u_{yy}=0$ and $v_{xx}+v_{yy}=0$). Suppose more that both $u$ and $v$ are not constant. Put $g = u_x - iu_y$ and $h = v_y + iv_x$. It is plain that $g$ and $h$ are holomorphic in $\Omega$ by checking their Cauchy-Riemann equations. If $Z(h) = \Omega$, then $v_x=v_y=0$ in $\Omega$. Let $D$ be any disc in $\Omega$ such that $\bar{D}\subset \Omega$. Then $v$ is a real part of a holomorphic function $V$ in $D$. By $V' = v_x - iv_y$, we have $V'=0$ in $D$, hence $V$ is constant in $D$ (for any $z,w$ in $D$, we have $V(z)-V(w) = \int_{[z,w]}V'(\zeta)\,d\zeta=0$), hence $v = \mathrm{const}$ in $D$. This argument shows that for any $z\in \Omega$, $v$ is constant in some neighborhood of $z$. Pick $z_0 \in \Omega$. Now, it is easy to see that the set $\{z\in \Omega: v(z) = v(z_0)\}$ is both closed and open in $\Omega$, hence is exactly $\Omega$ by the connectedness of $\Omega$. This means that $v$ is constant in $\Omega$, a contradiction with our assumption. So $Z(h)$ must have no limit point in $\Omega$. Put $k = g/h$, we have $k$ is a meromorphic function in $\Omega$. But $$ k = \frac{g}{h}=\frac{u_x - iu_y}{v_y+iv_x} = \frac{u_xv_y - u_yv_x}{v_x^2 + v_y^2}\in \mathbb{R}. \qquad (*) $$

There does not exist any nonempty open subset of $\mathbb{R}$. So $k$ is constant in each disc in which $g$ has no zero (by the open mapping theorem). Now, fix $a\in Z(h)$. Let $r>0$ such that $D(0;r)\subset \Omega$ and $h(z)\neq 0$ for every $z$ in $D'(0;r)$. Then we have $k$ is constant in each of four discs $D(a+r/2;r/2)$, $D(a+ir/2;r/2)$, $D(a-r/2;r/2)$, $D(a-ir/2;r/2)$. Because the first disc intersects with the second, the second intersects with the third, and the third intersects with the fourth, we conclude that $k$ is constant in the union of these four discs, hence is constant in $D(a;r/2)$ because $D(a;r/2)$ is a subset of this union. So $a$ is a removable singularity and $k$ is constant in the neighborhood $D(a;r/2)$ of $a$. By the connectedness of $\Omega$, we get $k$ is constant in $\Omega$. Denote this constant by $C$. By $(*)$, $C\in \mathbb{R}$, $C\neq 0$ (if not, $u_x=u_y=0$ in $\Omega$, hence $u$ is constant), and $u_x = Cv_y$, $u_y = -Cv_x$. We get the answer for the first question. Now, if $u^2$ is harmornic in $\Omega$, then, as above, we have $u$ is constant or $u_x = Cu_y$, $u_y = -Cu_x$ for some real constant $C\neq 0$. The second case leads to $u_x = u_y = 0$ in $\Omega$, which also implies $u$ is constant. We affirm the second statement. (A simpler way to get this affirmation is that we take the Laplacian of $u^2$ and get $u_x = u_y =0$.) For the last question, put $f = u + iv$. Because $f\in H(\Omega)$, we have $u$ and $v$ are harmonic in $\Omega$. Now, $|f|^2$ is harmonic iff $u^2 + v^2$ is harmonic. Take the Laplacian of $u^2+v^2$, we get $u_x = v_x = u_y = v_y =0$. So $u$ and $v$ are constant, hence $f$ is constant. $\Box$

**2.** Suppose $f$ is a complex function in a region $\Omega$, and both $f$ and $f^2$ are harmonic in $\Omega$. Prove that either $f$ or $\bar{f}$ is holomorphic in $\Omega$.

**Proof.** Put $f = u + iv$. Because $f$ is harmonic, $u$ and $v$ are harmonic. Suppose $f^2$ is harmonic in $\Omega$, which implies $u^2 - v^2$ and $uv$ are harmonic. By Exercise 1, $uv$ harmonic leads to $u$ is constant or $v$ is constant or there exists a nonzero real constant $C$ such that $u_x = Cv_y$ and $u_y = -Cv_x$. If $u$ is constant, then $v^2$ is harmonic (because $u^2-v^2$ is harmonic), hence $v$ is constant by Excercise 1. Similarly for the case $v$ is constant, we conclude than $f$ is constant if $u$ or $v$ is constant, hence both $f$ and $\bar{f}$ is holomorphic. Suppose $f$ is not constant. Take the Laplacian of $u^2-v^2$, we get $(u_x)^2+(u_y)^2 = (v_x)^2 + (v_y)^2$. Replace $u_x = Cv_y$ and $u_y = -Cv_x$ to this equation, we get $C^2 = 1$ unless $v_x = v_y = 0$ in $\Omega$. The case $C = 1$ leads to the Cauchy-Riemann equations of $f$, hence $f$ is holomorphic. The case $C=-1$ leads to the Cauchy-Riemann equations of $\bar{f}$, hence $\bar{f}$ is holomorphic. We get the conclusion. (There is another way which does not take the advantage of the first question in Excercise 1. Take the Laplacian of $u^2-v^2$ and $uv$, we get $(u_x)^2+(u_y)^2 = (v_x)^2 + (v_y)^2$ and $u_xv_x + u_yv_y=0$, which implies $(u_x+iv_x)^2 + (u_y+iv_y)^2 = 0$. Hence
$$
(u_x+v_y+iv_x - i u_y)(u_x -v_y+iv_x+iu_y) = 0.
$$

Use $ab=0 \Leftrightarrow a\bar{b}=0$, we get $$ (u_x+v_y+iv_x - i u_y)(u_x -v_y-iv_x-iu_y) = 0. $$

Put $g = u_x+v_y+iv_x - i u_y$ and $h = u_x -v_y-ivx-iu_y$, we get $gh=0$. It is easy to check that $g$ and $h$ are holomorphic (by checking the Cauchy-Riemann equations, we get $u_x-iu_y$ and $v_y+iv_x$ are holomorphic). We claim that $g=0$ in $\Omega$ or $h=0$ in $\Omega$. Suppose the contrary, because $Z(g)$ and $Z(h)$ are countable sets, $Z(gh)$ must be a countable set, which is a contradiction with $Z(gh)=\Omega$. The case $g=0$ leads to $f$ is holomorphic. The case $h=0$ leads to $\bar{f}$ is holomorphic. $\Box$

**3.** If $u$ is a harmonic function in a region $\Omega$, what can you say about the set of points at which the gradient of $u$ is $0$? (This is the set on which $u_x = u_y = 0$.)

**Proof.** $K = \{z:u_x(z) = u_y(z) = 0\}$ has no limit point in $\Omega$ (hence has at most countable elements) or $K = \Omega$.

Put $f= u_x - i u_y$. If $u$ is harmonic then $f$ is a holomorphic function by checking its Cauchy-Riemann equations. Moreover, because $K = Z(f)$, we get the conclusion. (The case $K = \Omega$ leads to $u_x = u_y = 0$ in $\Omega$. Hence $u$ is constant.) $\Box$

### Chapter 12 - The Maximum Modulus Principle¶

**1.** Suppose $\Delta$ is a closed equilaterial triangle in the plane, with vertices $a$, $b$, $c$. Find $\max(|z-a||z-b||z-c|)$ as $z$ range over $\Delta$.

**Proof.** $\max = \sqrt{3}|b-a|^3/8$.

By the maximum modulus theorem, we have $$ \max\limits_{z\in\Delta}(|(z-a)(z-b)(z-c)|) = \max\limits_{z\in\partial\Delta}(|(z-a)(z-b)(z-c)|). $$

Now, it is elementary to find the maximum of the right hand side, which is attained when $z$ is one of the middle points of the edges $ab$, $bc$, and $ca$. Indeed, without loss of generality, suppose the maximum is attained at $z$ on $ab$. Put $x = |z-a|$ and $l = |b-a|$. So the maximum is $$ x(l-x)\sqrt{(x-l/2)^2 + (\sqrt{3}l/2)^2} = x(l-x)\sqrt{l^2-x(l-x)}. $$

When $x$ ranges over $[0,l]$, $x(l-x)$ ranges over $[0,l^2/4]$. Put $t = x(l-x)$, we get the maximum is $t\sqrt{l^2-t}$, or $\sqrt{l^2t^2-t^3}$. The derivative of $l^2t^2-t^3$ according to $t$ is $(2l^2-3t)t$, which is greater than or equal to $0$ when $t$ ranges over $[0,l^2/4]$. So $t\sqrt{l^2-t}$ is increasing when $t$ runs from $0$ to $l^2/4$, and the maximum is attained when $t = l^2/4$, which is the case $x = l/2$. $\Box$

**3.** Suppose $f\in H(\Omega)$. Under what conditions can $|f|$ have a local minimum in $\Omega$.

**Proof.** $f$ is constant or $f$ has at least one zero in $\Omega$.

On the contrary, suppose $f$ is not constant and $f$ has no zero in $\Omega$. So $f^{-1}\in H(\Omega)$ and $|f|$ has a local minimum at $a$ iff $|f^{-1}|$ has a local maximum at $a$. By applying the maximum modulus theorem, if $|f^{-1}|$ has a local maximum then $f^{-1}$ must be constant, hence $f$ is constant. So $f$ has no local minimum on $\Omega$. (For more details, we consider the case $f$ is not constant. Clearly $|f|$ has local minima at zeroes of $f$. Reversely, let $a$ be a point at which $f$ has a local minimum. If $a \neq 0$, then $f\neq 0$ in an open neighborhood of $a$. As above, $f$ must be constant in this neighborhood, hence be constant in $\Omega$. So, in this case, we can see that the set of all local minima of $f$ is $Z(f)$, the set of all zeroes of $f$.) $\Box$

**4.** $(a)$ Suppose $\Omega$ is a region, $D$ is a disc, $\bar{D}\subset \Omega$, $f\in H(\Omega)$, $f$ is not constant, and $|f|$ is constant on the boundary of $D$. Prove that $f$ has at least one zero in $D$.

$(b)$ Find all entire functions $f$ such that $|f(z)| = 1$ whenever $|z|=1$.

**Proof.** $(a)$ Suppose that $f$ has no zero in $D$. If $|f|=0$ on $\partial D$, then by the maximum modulus theorem, $|f|=0$ in $D$, hence $f$ is constant. So $f\neq 0$ on $\partial D$, hence $f \neq 0$ on $\bar{D}$. Put $c>0$ be the constant value of $|f|$ on $\partial D$. By the maximum modulus theorem, we have $|f|\leq c$ in $D$. Consider the function $f^{-1}$ on $\bar{D}$. It is clear that $f^{-1}\in C(\bar{D})$ and $f^{-1}\in H(D)$. By the maximum modulus theorem, we have $|f^{-1}| \leq 1/c$ in $D$, hence $|f| \geq c$ in $D$. So $|f| = c$ in $D$, hence $f(D) \subset \partial D(0;c)$. By the open mapping theorem, $f$ must be constant in $D$, hence be constant in $\Omega$. This contradiction shows that $f$ must have at least one zero in $D$. (In this proof, we just need $f$ is holomorphic in $D$ and continuous on $\bar{D}$.)

$(b)$ $f$ is of the form $cz^m$, where $|c| = 1$ and $m \geq 0$.

Suppose that $f$ is not constant. By $(a)$, $f$ has at least one zero in $D(0;1)$. The set of all zeroes of $f$ has no limit point in $\mathbb{C}$. Because $\bar{D}(0;1)$ is compact, the number of zeroes of $f$ in $D(0;1)$ is finite. Let $a_1,a_2,\ldots,a_n\in D(0;1)$ be such zeroes and $m_1,m_2,\ldots,m_n$ be the orders of zeroes of $f$ at these points. Put $$\begin{aligned} g(z) &= f(z)\prod\limits_{i=1}^n \left(\frac{1-\bar{a}_iz}{z-a_i}\right)^{m_i}\\ &=\frac{f(z)}{(z-\bar{a}_1)^{m_1}(z-\bar{a}_2)^{m_2}\ldots(z-\bar{a}_n)^{m_n}}\prod\limits_{i=1}^n (1-\bar{a}_iz) \end{aligned} $$

for all $z\in \mathbb{C}$. It is clear that $g\in H(\mathbb{C})$ and $g$ has no zero in $D(0;1)$. Moreover, for $|z| = 1$, we have $$ \left|\frac{1-\bar{a}_iz}{z-a_i}\right| = \left|z.\frac{1/z-\bar{a}_i}{z-a_i} \right| = \left|z.\frac{\bar{z}-\bar{a}_i}{z-a_i} \right| = \left|z.\frac{\overline{z - a_i}}{z-a_i} \right| = 1\qquad (i=\overline{1,n}). $$

Hence $|g(z)|=|f(z)|=1$ for every $z\in \partial D(0;1)$. By $(a)$, $g$ must be a constant, which will be denoted by $c$. Clearly $|c|=1$. If there is exists $i$ such that $a_i\neq 0$, then $g(1/\bar{a}_i) = 0$, which is a contradiction to $|c|=1$. So $n=1$ and $a_1 = 0$. This means that $f(z) = cz^{m_1}$, where $|c|=1$ and $m_1 > 0$. Together with the case $f$ is a constant, we see that $f$ must be in the form $cz^{m}$, where $|c|=1$ and $m\geq 0$. In the end, it is clear that these functions satisfy our requirement. $\Box$

**5.** Suppose $\Omega$ is a bounded region, $\{f_n\}$ is a sequence of continuous functions on $\bar{\Omega}$ which are holomorphic in $\Omega$, and $\{f_n\}$ converges uniformly on the boundary of $\Omega$. Prove that $\{f_n\}$ converges uniformly on $\bar{\Omega}$.

**Proof.** Fix $\epsilon > 0$. There is $N$ large enough such that for every $m,n > N$, we have $|f_n(z) - f_m(z)| < \epsilon$ for every $z\in \partial \Omega$. By the maximum modulus theorem, for every $m,n>N$, we have
$$
|f_n(z) - f_m(z)| < \epsilon \qquad (*)
$$

for every $z\in \bar{\Omega}$. So for each $z\in \partial\Omega$, $\{f_n(z)\}$ is a Cauchy sequence, hence converges. For each $z$, put $f(z)$ be this limit. Let $m$ converges to $\infty$ in $(*)$, we get the conclusion that $\{f_n\}$ converges to $f$ uniformly on $\bar{\Omega}$. $\Box$

**6.** Suppose $f\in H(\Omega)$, $\Gamma$ is a cycle in $\Omega$ such that $\mathrm{Ind}_{\Gamma}(\alpha) = 0$ for all $\alpha \notin \Omega$, $|f(\zeta)|\leq 1$ for every $\zeta \in\Gamma^*$, and $\mathrm{Ind}_{\Gamma}(z) \neq 0$. Prove that $|f(z)|\leq 1$.

**Proof.** Let $U$ be an open connected component of $\mathbb{C}\backslash \Gamma$ such that the index of any point in $U$ with respect to $\Gamma$ is not zero. So $U$ is bounded and, by assumption, $U \subset \Omega$. Let $z \in \partial U$. Because $z \notin U$ and also $z$ can not be in any other component of $\mathbb{C}\backslash \Gamma$, we have $z\in \Gamma$. So $\partial U\subset \Gamma$ and $\bar{U} \subset \Omega$. By assumption, we get $\|f\|_{\partial U} \leq 1$. Moreover, because $U$ is bounded, we have $\bar{U}$ is compact. By the maximum modulus theorem, we get $|f(z)| \leq 1$ for every $z\in U$. So $|f(z)|\leq 1$ for every $z$ such that $\mathrm{Ind}(z) \neq 0$. $\Box$

**7.** In the proof of Theorem 12.8 it was tacitly assumed that $M(a)>0$ and $M(b) > 0$. Show that the theorem is true if $M(a) = 0$, and that then $f(z) = 0$ for all $z\in \Omega$.

**Proof.** When $M(a) = 0$, consider the segment $L = \{a+ yi: y\in (0,1)\}$, it is plain that $f$ is vanishing on $L$. Put $\Theta = \Omega \cup L \cup \{x+yi: 2a - b < x < b\}$. By the Schwarz reflection principle, we get a function $F \in H(\Theta)$ such that $F=f$ on $\Omega\cup L$. Because $\Theta$ is a region and $F$ is vanishing on $L$, $F$ must be vanishing on $\Theta$, hence $f$ is vanishing on $\Omega$. In other words, we have $M(x) = 0$ for all $x \in (a,b)$. (To verify that we can apply the Schwarz reflection principle here, consider the map $z \to i(z-a)$.) $\Box$

**15.** Suppose $f\in H(U)$. Prove that there is a sequence $\{z_n\}$ in $U$ such that $|z_n|\to 1$ and $\{f(z_n)\}$ is bounded.

**Proof.** The problem is plain for the case $f$ is constant. For the case $f$ has no zero in $U$. For each $r\in (0,1)$, apply the maximum modulus theorem for the function $f^{-1}$ with the domain $\bar{D}(0;r)$, we get $|f(0)|\geq \min_{z\in\partial D(0;r)}|f(z)|$, hence we get a point $a_r$ such that $|a_r| = r$ and $|f(a_r)|\leq |f(0)|$. So the sequence $\{a_{1-1/n}\}$ satisfies our requirement. Now suppose that $f$ is not constant and $f$ has at least one zero in $U$. Put $Z(f)$ be the set of zeroes of $f$ in $U$. It is clear that $Z(f)$ has no limit point in $U$. If $Z(f)$ is an infinite set then it must have a limit point in $\bar{U}$, which is actually on $\partial U$. So we obtain a sequence $\{z_n\}\subset U$ converging to this limit point (hence $|z_n|\to 1$) and $|f(z_n)|=0$ for every $n$. We get the conclusion for this case. The remaining case is that $Z(f)$ is a finite set. Put $Z(f) = \{a_1,a_2,\ldots,a_n\}$ and let $m_1,m_2,\ldots,m_n$ be the orders of zeroes of $f$ at $a_1,a_2,\ldots,a_n$ correspondingly. Put
$$
g(z) = \frac{f(z)}{(z-a_1)^{m_1}(z-a_2)^{m_2}\ldots(z-a_n)^{m_n}}\qquad (z\in U).
$$

It is clear that $g\in H(U)$ and $g$ has no zero in $U$. As above, we get a sequence $\{z_n\}$ in $U$ such that $|z_n|\to 1$ and $|g(z_n)| < M$ for some $M > 0$. For $|z| > \max\{|a_1|,|a_2|,\ldots,|a_n|\}$, we have $$\begin{aligned} |f(z)| &= |g(z)||z_n-a_1|^{m_1}|z_n-a_2|^{m_2}\ldots|z-a_n|^{m_n}\\ &\leq |g(z)|(1-|a_1|)^{m_1}(1-|a_2|)^{m_2}\ldots(1-|a_n|)^{m_n}. \end{aligned} $$

So for $n$ large enough, we have $|z_n| > \max\{|a_1|,|a_2|,\ldots,|a_n|\}$ and $$ |f(z_n)| < M(1-|a_1|)^{m_1}(1-|a_2|)^{m_2}\ldots(1-|a_n|)^{m_n}. $$

The sequence $\{z_n\}$ is the one we want to find. $\Box$

**16.** Suppose $\Omega$ is a bounded region, $f\in H(\Omega)$, and
$$
\limsup\limits_{n\to\infty} |f(z_n)| \leq M
$$

for every sequence $\{z_n\}$ in $\Omega$ which converges to a boundary point of $\Omega$. Prove that $|f(z)|\leq M$ for all $z\in \Omega$.

**Proof.** It is enough to prove that $|f(z)| \leq M + \epsilon$ for every $z\in \Omega$ and every $\epsilon > 0$. Instead, fix $\epsilon >0$, we will prove that the set $A = \{z\in \Omega: |f(z)| > M+\epsilon\}$ is empty. On the contrary, suppose that $A$ is not empty. It is plain that $A$ is open. Let $U$ be a nonempty connected open component of $A$ and let $z\in \partial U\subset \bar{\Omega}$. There is a sequence $\{z_n\}\subset U$ such that $z_n \to z$ and $|f(z_n)|> M+\epsilon$ for all $n$. Then we have $\limsup_{n\to\infty} |f(z_n)| \geq M + \epsilon$, which is a contradiction to our hypothesis if $z\in \partial \Omega$. So $z\in \Omega$ for all $z\in\partial U$. In other words, $\bar{U}\subset \Omega$. Because $\Omega$ is bounded, we have $\bar{U}$ is a compact set. Moreover, for $z\in \partial U$, we have $|f(z)| \geq M + \epsilon$, hence $|f(z)| = M +\epsilon$ (if not then $z\in A$, hence $z\in A\backslash U$, which is an open set, hence $z\notin \bar{U}$). Apply the maximum modulus theorem for $f$ with the domain $\bar{U}$, we get $|f(z)|\leq M + \epsilon$ for every $z\in U$, which is a contradiction with our definition of $U$. So $A$ must be empty. $\Box$

### Chapter 13 - Approximation by Rational Functions¶

**1.** Prove that every meromorphic function on $S^2$ is rational.

**Proof.** Note that $f$ must have an isolated singularity at $\infty$, which means that $f$ is holomorphic in $D'(\infty;r)=\{z\in \mathbb{C}:|z|> r\}$. This implies that the set of all poles of $f$ in $\mathbb{C}$ is finite (because this set has no limit point and it is a subset of the compact set $\overline{D}(0;r)$). If we subtract from $f$ the sum of all principle parts at these poles, we get an entire function $g$ with an isolated singularity at $\infty$. Of course, this singularity of $g$ at $\infty$ can not be an essential singularity because of our hypothesis that $f$ is meromorphic on $S^2$. If this singularity is a removable singularity then $g$ is bounded in a neighborhood of $\infty$, hence is bounded in $\mathbb{C}$, hence is constant by Liouville's theorem. If this singularity is a pole, the principal part of $g$ at $\infty$ is a polynomial. Subtract this polynomial from $g$, we again get an entire function with a removable singularity at $\infty$. Both cases lead to the same conclusion that $f$ is a rational function. $\Box$

**2.** Let $\Omega = \{z:|z|<1 \text{ and }|2z-1|>1\}$, and suppose $f\in H(\Omega)$.

$(a)$ Must there exist such a sequence of polynomials $P_n$ such that $P_n \to f$ uniformly on compact subsets of $\Omega$?

$(b)$ Must there exist such a sequence which converges to $f$ uniformly in $\Omega$?

$(c)$ Is the answer to $(b)$ changed if we require more of $f$, namely, that $f$ be holomorphic in some open set which contains the closure of $\Omega$?

**Proof.** $(a)$ Yes. Because $S^2\backslash \Omega$ is connected (then apply Runge's theorem).

$(b)$ No. See $(c)$ or for simple, consider the function $f(z) = z^{-1}$ with a similar argument.

$(c)$ No. Consider the function $f(z) = (z-\frac{1}{10})^{-1}$. Suppose that there exists a sequence of polynomials $P_n$ such that $P_n$ converges to $f$ uniformly in $\Omega$. This implies there there exists a polynomial $P$ such that $$ |P(z) - (z-1/10)^{-1}| < 1 $$ for every $z\in \Omega$. Fix $z \in \partial D(0;1)$. Let $\{z_n\}$ be a sequence in $\Omega$ converges to $z$. We have $$\begin{aligned} |P(z)| = \lim\limits_{n\to\infty}|P(z_n)| &\leq 1 + \lim\limits_{n\to\infty}\frac{1}{\left| z_n-1/10\right| } = 1 + \frac{1}{\left| z-1/10\right| }\\ &\leq 1 + \frac{1}{1 - 1/10} < 3. \end{aligned} $$

By the maximum modulus theorem, $|P(z)| < 3$ for every $z \in D(0;1)$. However, for $z = \frac{-1}{10}$, we have $$ |P(-1/10)| > \frac{1}{\left|-1/10 - 1/10\right|} - 1 = 4. $$

This contradiction shows that there is not any sequence of polynomials which converges to $(z-\frac{1}{10})^{-1}$ uniformly in $\Omega$. $\Box$

**3.** Is there a sequence of polynomials $P_n$ such that $P_n(0)=1$ for $n=1,2,3,\ldots$, but $P_n(z)\to 0$ for every $z\neq 0$, as $n\to \infty$?

**Proof.** Yes. For each $n$, let $K_n$ be the compact set $D_{n} \cup L_{n} \cup \{0\}$, where $D_{n} = \{z\in \bar{D}(0;n):|\mathrm{Im} z|\geq \frac{1}{n}\}$, $L_n = \{x+yi: y = 0, \frac{1}{n}\leq |x|\leq n\}$. Let $U_n$ be an neighborhood of $D_n \cup L_n$ and $V_n$ be an neighborhood of $0$ such that $U_n$ and $V_n$ are disjoint. Let $f_n \in H(U_n \cup V_n)$ be a function whose values are $0$ in $U_n$ and $1$ in $V_n$. Because $S^2\backslash K_n$ is connected, apply Runge's theorem, we get a polynomial $P_n$ such that $|P_n(z) - f_n(z)| < \frac{1}{n}$ for every $z\in K_n$, hence $|P_n(z)| < \frac{1}{n}$ for every $z\in D_n \cup L_n$ and $|P_n(0)-1| < \frac{1}{n}$. Now, because $\{D_n\cup L_n\}$ is an increasing sequence of sets whose union is $\mathbb{C}\backslash \{0\}$, we can see that the sequence of polynomials $P_n$ satisfies $P_n(0) \to 1$ as $n \to \infty$ and for $z \neq 0$, $P_n(z) \to 0$ as $n \to \infty$. Replace $P_n$ by $P_n - P_n(0) + 1$, we get a sequence of polynomials satisfying our requirement. $\Box$

**4.** Is there a sequence of polynomials $P_n$ such that
$$
\lim\limits_{n\to\infty}P_n(z) =
\begin{cases}
1 & \text{if }\mathrm{Im}z > 0,\\
0 &\text{if } z \text{ is real},\\
-1 & \text{if }\mathrm{Im}z < 0?
\end{cases}
$$

**Proof.** Yes. For each $n$, let $K_n$ be the compact set $D_n \cup E_n \cup L_n $, where $D_n = \{z \in\bar{D}(0;n):\mathrm{Im}z\geq \frac{1}{n}\}$, $E_n = \{z \in\bar{D}(0;n):\mathrm{Im}z\leq -\frac{1}{n}\}$, and $L_n = \{ x + yi:y = 0, |x| \leq n\}$. Let $U_n$ be an neighborhood of $D_n$, $V_n$ be an neighborhood of $E_n$, and $W_n$ be an neighborhood of $L_n$ such that $U_n$, $V_n$, and $W_n$ are pairwise disjoint. Now let $f_n\in H(U_n\cup V_n \cup W_n)$ be a function whose values are $1$ in $U_n$, $-1$ in $V_n$, and $0$ in $L_n$. Because $S^2\backslash K_n$ is connected, apply Runge's theorem, we get a polynomial $P_n$ such that $|P_n(z)-f_n(z)| < \frac{1}{n}$ for every $z\in K_n$, hence $|P_n(z)-1| < \frac{1}{n}$ for every $z\in D_n$, $|P_n(z)+1| < \frac{1}{n}$ for every $z\in E_n$, and $|P_n(z)| < \frac{1}{n}$ for every $z\in L_n$. Because $\{D_n\}$, $\{E_n\}$, and $\{L_n\}$ are three increasing sequences of sets whose unions are open upper half plane, open lower half plane, and real line correspondingly, we can see that the sequence of polynomials $P_n$ satisfies our requiment. $\Box$

**7.** Show that in Theorem 13.9 we need not assume that $A$ intersects each component of $S^2\backslash \Omega$. It is enough to assume that the closure of $A$ intersects each component of $S^2\backslash \Omega$.

**Proof.** We have $\Omega$ is the union of a sequence $\{K_n\}$ of compact sets such that $K_n$ lies in the interior of $K_{n+1}$ for every $n$, every compact subsets of $\Omega$ lies in some $K_n$, and for every $n$, every component of $S^2\backslash K_n$ contains a component of $S^2 \backslash \Omega$ (Theorem 13.3). Fix $n$. Follow the proof of Theorem 13.9, it is enough to show that each component of $S^2 \backslash K_n$ contains a point of $A$. Indeed, let $U$ be a component of $S^2\backslash K_n$. It is clear that $U$ is open. Moreover, $\bar{A}$ intersects each component of $S^2\backslash \Omega$, hence intersects $U$. So $U \cap A \neq \varnothing$. $\Box$

**8.** Prove the Mittag-Lefler theorem for the case in which $\Omega$ is the whole plane, by a direct argument which makes no appeal to Runge's theorem.

**Proof.** For each $n$, denote $\bar{D}(0;n)$ by $K_n$ for simple. Put $A_1 = A \cap K_1$, and for each $n \geq 2$, put $A_n = A\cap(K_n\backslash K_{n-1})$. It is easy to see that each $A_n$ is finite. Now, for each $n$, put
$$
Q_n(z) = \sum\limits_{\alpha \in A_n}P_{\alpha}(z).
$$

For each $n$, we have $Q_n$ is a rational function and the poles of $Q_n$ lie in $A_n$. Fix $n \geq 2$. We have $Q_n$ is holomorphic in an open set containing $K_{n-1}$. Consider the power series representation of $Q_n$ at $0$ and recall that $K_{n-1}$ is $\bar{D}(0;n-1)$, we get the power series converges to $Q_n$ uniformly in $K_{n-1}$. This implies that we have a polynomial $R_n$ such that $$ |R_n(z) - Q_n(z)| < 2^{-n}\qquad (z\in K_{n-1}). $$

Follow the proof of the Mittag-Leffler theorem, we can see that the function $$ f(z) = Q_1(z) + \sum_{n=2}^{\infty}(Q_n(z) - R_n(z))\qquad (z\in \mathbb{C}) $$ has the desired properties. $\Box$

**9.** Suppose $\Omega$ is a simply connected region, $f\in H(\Omega)$, $f$ has no zero in $\Omega$, and $n$ is a positive integer. Prove that there exists a $g\in H(\Omega)$ such that $g^n = f$.

**Proof.** Let $h$ be a holomorphic logarithm of $f$ in $\Omega$ (the existence comes from the hypotheses that $\Omega$ is a simply connected region, $f\in H(\Omega)$, and $f$ has no zero in $\Omega$). Now, the function $g = \exp(h/n)$ is in $H(\Omega)$ and $g^n = \exp(h) = f$. $\Box$

**10.** Suppose $\Omega$ is a region, $f\in H(\Omega)$, and $f\neq 0$. Prove that $f$ has a holomorphic logarithm in $\Omega$ if and only if $f$ has holomorphic $n$-th roots in $\Omega$ for every positive integer $n$.

**Proof.** The $(\Rightarrow)$ side is clear, an $n$-th root of $f$ is $\exp(g/n)$ where $g$ be a holomorphic logarithm in $\Omega$ of $f$. For the $(\Leftarrow)$ side, at first, we will show that $f$ has no zero in $\Omega$. Indeed, suppose $f(a) = 0$ for some $a\in \Omega$. Because $f\neq 0$, we have $f(z) = (z-a)^m g(z)$, where $m>0$ is the order of zero of $f$ at $a$, $g\in H(\Omega)$, and $g(a) \neq 0$. Consider an holomorphic $(m+1)$-th root $\varphi$ of $f$. Because $\varphi(a) = 0$ and $\varphi \neq 0$, we have $\varphi(z) = (z-a)h(z)$, where $h\in H(\Omega)$. Therefore $g(z) = (z-a)h(z)^{m+1}$ and this gives us $g(a) = 0$ (a contradiction). To show that $f$ has a holomorphic logarithm in $\Omega$, it is enough to show that
$$
\int_{\gamma}\frac{f'(z)}{f(z)}\,dz = 0\qquad (*)
$$
for every closed path $\gamma$ in $\Omega$. Indeed, suppose $(*)$ is true for every closed path in $\Omega$. Fix $z_0\in \Omega$ and let $l_0$ be a logarithm of $f(z_0)$ (this logarithm exists because $f(z_0)\neq 0$). Let $L$ be a function in $\Omega$ defined by
$$
L(z) = l_0 + \int_{\Gamma(z)}\frac{f'(\zeta)}{f(\zeta)}\,d\zeta\qquad (z\in \Omega),
$$
where $\Gamma(z)$ is any path in $\Omega$ from $z_0$ to $z$. Because of $(*)$, this function is well-defined. Fix $a\in \Omega$. For any $D(a;r)\subset \Omega$ and any $z\in D'(a;r)$, we have
$$
\frac{L(z)-L(a)}{z-a}-\frac{f'(a)}{f(a)} = \frac{1}{z-a}\int_{[a,z]}\left(\frac{f'(\zeta)}{f(\zeta)}-\frac{f'(a)}{f(a)}\right)\,dz.
$$

Because $f'/f$ is continuous at $a$, it is easy to see that $L'(a) = f'(a)/f(a)$. So $L\in H(\Omega)$ and the formula $L'=f'/f$ gives us $(\exp(L)/f)'=0$. Hence $\exp(L)/f = \mathrm{const} = \exp(L(z_0))/f(z_0)= 1$. So $L$ is a holomorphic logarithm of $f$ in $\Omega$. In the end, to show that $(*)$ is true, we notice that $\frac{1}{2\pi i}$ times the left hand side of $(*)$ is the index of $f\circ \gamma$ at $0$, which is an integer number. Moreover, if $g$ is an holomorphic $n$-th root of $f$ in $\Omega$, then $$ \frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}\,dz = \frac{n}{2\pi i}\int_{\gamma}\frac{g'(z)}{g(z)}\,dz. $$

The right hand side is an integer which is divisible by $n$. We get the conclusion by noticing that $0$ is the unique integer number which is divisible by all positive integer. $\Box$

**11.** Suppose that $f_n \in H(\Omega)$ ($n=1,2,3,\ldots$), $f$ is a complex function in $\Omega$, and $f(z) = \lim_{n\to\infty}f_n(z)$ for every $z\in \Omega$. Prove that $\Omega$ has a dense open subset $V$ on which $f$ is holomorphic.*Hint.* Put $\varphi = \sup|f_n|$. Use Baire's theorem to prove that every disc in $\Omega$ contains a disc on which $\varphi$ is bounded. Apply Excercise 10.5.

(In general, $V\neq \Omega$. Compare Excercises 3 and 4.)

**Proof.** Put $\varphi(z) = \sup_n |f_n(z)|$ for every $z \in \Omega$. It is well-defined because $\lim_{n\to\infty} f_n(z)$ exists (and equals to $f(z)$) for every $z \in \Omega$. Fix $B$ be a closed disc in $\Omega$ (so $B$ is complete). For each $n$, put $A_n = \{z\in B:\varphi(z) \leq n\}$. By the definition of $\varphi$, it is clear that
$$
A_n = \bigcap\limits_{m=1}^{\infty} \{z\in B: |f_m(z)|\leq n\}.
$$

Hence $A_n$ is a closed set in $B$ for each $n$. By Baire's theorem, there exists an $n$ such that $A_n$ is not a nowhere-dense subset of $B$. So $A_n$ contains an open disc $U_B\subset B$ in which $\varphi$ is bounded from above by $n$. Now, apply Exercise 10.5, we get $f \in H(U_B)$. Let $$ V = \bigcup\limits_B U_B, $$ where $B$ ranges over all closed disc in $\Omega$. It is clear that $V$ is a dense open subset of $\Omega$ and $f\in H(V)$. $\Box$

### Chapter 14 - Conformal Mapping¶

**10.** Suppose $f$ and $g$ are holomorphic mappings of $U$ into $\Omega$, $f$ is one-to-one, $f(U) = \Omega$, and $f(0) = g(0)$. Prove that
$$
g(D(0;r))\subset f(D(0;r))\qquad (0< r < 1).
$$

**Proof.** Put $h=f^{-1}\circ g$. Then $h(U)\subset U$ and $h(0) = 0$. By Schwartz's lemma, we have $|h(z)|\leq |z|$ for all $z\in U$. Fix $r\in (0,1)$. If $z\in D(0;r)$ then also $h(z)\in D(0;r)$, hence $f^{-1}\circ g (z) \in D(0;r)$, and hence $g(z) \in f(D(0;r))$. So $g(D(0;r))\subset f(D(0;r))$. $\Box$

**12.** Suppose $\Omega$ is a convex region, $f\in H(\Omega)$, and $\mathrm{Re}f'(z)>0$ for all $z\in \Omega$. Prove that $f$ is one-to-one in $\Omega$. Is the result changed if the hypothesis is weakened to $\mathrm{Re}f'(z) \geq 0$? (Exclude the trivial case $f=$ constant.) Show by an example that "convex" cannot be replaced by "simply connected."

**Proof.** No. $\Omega = \mathbb{C}\backslash [0,\infty)$, $f(z) = -iz^{3/2}+z/\sqrt{2}$.

Let $a,b\in \Omega$ such that $a \neq b$. We have $$ f(b) - f(a) = \int_{[a,b]}f'(z)\,dz = (b-a)\int_0^1 f'(a+(b-a)t)\,dt. $$

Hence $$\begin{aligned} \mathrm{Re}\left[\frac{f(b)-f(a)}{b-a}\right] &= \mathrm{Re}\left[\int_0^1 f'(a+(b-a)t)\,dt\right]\\ &= \int_0^1 \mathrm{Re}\left[f'(a+(b-a)t)\right]\,dt > 0. \end{aligned} $$

So we must have $f(b) \neq f(a)$, and this implies that $f$ is one-to-one. Now suppose that $\mathrm{Re}f'(z)\geq 0$ for all $z\in \Omega$. Let $Z = \{z\in \Omega: \mathrm{Re}f'(z) = 0\}$. Of course, $Z$ is a closed set. By the mean value property of $\mathrm{Re}f'$ and $\mathrm{Re}f'(z)\geq 0$ for all $z\in \Omega$, it is easy to show that $Z$ is an open set. So $Z = \varnothing$ or $Z = \Omega$. The first case implies, as we have shown, $f$ is one-to-one in $\Omega$. The second case implies $f'(\Omega) \subset i\mathbb{R}$. So, by the open mapping theorem, $f'$ is constant in $\Omega$, hence $f$ is constant or linear. The linear case also leads to $f$ is one-to-one. So, unless $f$ is constant, $f$ is always one-to-one when $\mathrm{Re}f'(z)\geq 0$ for all $z\in \Omega$. For an example that "convex" cannot be replaced by "simply connected", consider $\Omega = \mathbb{C}\backslash [0,\infty)$ and $f(z) = -iz^{3/2}+z/\sqrt{2}$ with the branch cut for $z^{1/2}$ is $[0,\infty)$. It is easy to check that $f'(z) = -\frac{3i}{2}z^{1/2} + 1/\sqrt{2}$ has a positive real part and $f(i) = f(-i)$. $\Box$

**13.** Suppose $\Omega$ is a region, $f_n\in H(\Omega)$ for $n=1,2,3,\ldots$, each $f_n$ is one-to-one in $\Omega$, and $f_n\to f$ uniformly on compact subsets of $\Omega$. Prove that $f$ is either constant or one-to-one in $\Omega$. Show that both cases can occur.

**Proof.** Suppose there exists $a,b\in \Omega$ such that $a\neq b$ and $f(a) = f(b)$. Choose $r > 0$ such that $\bar{D}(a;r)\subset \Omega$ and $b \notin D(a;r)$. By the assumption, we have $f_n(z) - f_n(b) \to f(z) -f(b)$ uniformly on $\bar{D}(a;r)$. Now by Exercise 10.20 with a note that $f_n(z) - f_n(b) \neq 0$ for all $z\in D(a;r)$ and all $n$, we have $f(z) - f(b)\neq 0$ for all $z\in D(a;r)$ or $f(z) -f(b)=0$ for all $z\in D(a;r)$. The first case cannot happen because $f(a) = f(b)$. And the second case implies that $f$ is constant. $\Box$

**16.** Let $\mathcal{F}$ be the class of all $f\in H(U)$ for which
$$
\iint\limits_U |f(z)|^2 \, d x\,d y \leq 1.
$$

Is this a normal family?

**Proof.** Let $a \in U$ and $R > 0$ such that $D(a;2R) \subset U$. For $z\in D(a;R)$, we have $\bar{D}(z;r) \subset U$ for all $r\in (0,R]$. By the mean value property of $f$, we have
$$
f(z) = \frac{1}{2\pi}\int_{0}^{2\pi} f(z + re^{it})\, dt.
$$

Multiply two sides with $r$ and integrate them according to $r$ from $0$ to $R$, we get $$ \frac{R^2}{2}f(z) = \frac{1}{2\pi}\int_0^R \int_0^{2\pi}rf(z + re^{it})\,dt. $$

Hence, by Hölder's inequality, we have $$ \begin{aligned} |f(z)| &\leq \frac{1}{\pi R^2}\left[\int_0^R \int_0^{2\pi} r\,dr\,dt\right]^{1/2}\left[\int_0^R \int_0^{2\pi} r|f(z + re^{it})|^2\,dr\,dt\right]^{1/2}\\ &= \frac{1}{\sqrt{\pi} R}\left[\iint\limits_{D(z;R)}|f(w)|\,dw\right]^{1/2} \leq \frac{1}{\sqrt{\pi} R}. \end{aligned} $$

For each compact set $K \subset \Omega$, we can cover it by finite disc $D(a;R)$ with the property that $D(a;2R) \subset U$. By the above argument, $\mathcal{F}$ is uniformly bounded in each disc, hence is uniformly bounded on $K$. By Theorem 14.6, we conclude that $\mathcal{F}$ is a normal family. $\Box$

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