# Examination 6¶

## 2004 Apr¶

Instructions. Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well written solutions will count more than several partial solutions.

Notation. $$D(z_0,R) = \{z\in \mathbb{C}: |z-z_0|<R\}$$ $$R>0$$. For an open set $$G\subseteq \mathbb{C}$$, $$H(G)$$ will denote the set of functions which are analytic in $$G$$.

Problem 34

Let $$\gamma$$ be a rectifiable curve and let $$\varphi \in C(\gamma^*)$$. (That is, $$\varphi$$ is a continuous complex function defined on the trace, $$\gamma^*$$, of $$\gamma$$.)

Let $$F(z) = \int_\gamma \frac{\varphi(\omega)}{(\omega-z)} \, d\omega, \quad z\in \mathbb{C}\setminus \gamma^*$$.

Prove that $$F'(z) = \int_\gamma \frac{\varphi(\omega)}{(\omega-z)^2} \, dw, \quad z\in \mathbb{C}\setminus \gamma^*$$, without using Leibniz’s Rule.

Problem 35
1. State the Casorati-Weierstrass theorem.

2. Evaluate the integral

$I =\frac{1}{2\pi i} \int_{|z|=R} (z-3) \sin\left(\frac{1}{z+2}\right)\, dz \; \text{ where } R\geq 4.$
Problem 36

Let $$f(z)$$ be an entire function such that $$f(0)=1$$, $$f'(0)=0$$ and

(17)$0<|f(z)|\leq e^{|z|}, \text{ for all } z\in \mathbb C.$

Prove that $$f(z)=1$$ for all $$z\in \mathbb{C}$$.

Problem 37

Let $$C$$ be an arbitrary circle through $$-1$$ and $$1$$. Suppose that $$z_1$$ and $$z_2$$ are two points which do lie on the circle $$C$$ and satisfy $$z_1z_2 = 1$$. Show that one of these points lies inside $$C$$ and the other lies outside $$C$$.

Problem 38

Show that there is no one-to-one analytic function that maps $$G = \{z : 0 < |z| < 1\}$$ onto the annulus $$\Omega = \{z : r < |z| < R\}$$, where $$r>0$$.

Problem 39
1. State a theorem that gives a sufficient condition for a family $$\mathcal F$$ of analytic functions to be normal in a domain $$G$$.

2. Let $$\mathcal F\subseteq H(D)$$ be a family of analytic functions on the open unit disk $$D = D(0,1)$$. Let $$\{M_n\}$$ be a sequence of positive real numbers such that $$\varlimsup_{n\to \infty} \sqrt[n]{M_n} < 1$$. If for each $$f(z) = \sum_{n=0}^\infty a_n z^n \in \mathcal F$$, $$|a_n| \leq M_n$$ for all $$n$$, prove that $$\mathcal F$$ is a normal family.

Problem 40

Is there a harmonic function $$u(z)$$ defined on the open unit disk, $$D(0, 1)$$, such that $$u(z_n) \to \infty$$ whenever $$|z_n|\to 1^-$$? Prove your answer.

Problem 41

Let $$G$$ be a simply connected domain with at least 2 boundary points. Let

$S = \{\psi \in H(G) \mid \psi\colon G \to D(0,1), \psi \text{ is one-to-one} \}.$

Prove, without using the Riemann mapping theorem, that the set $$S$$ is nonempty.

## Solutions¶

Solution to Problem 34

We give two alternative proofs of this result.

Proof 1. Consider, for some $$z_0\in \mathbb C \setminus \gamma^*$$,

$\begin{split}\frac{1}{\omega - z} &= \frac{1}{(\omega - z_0) - (z - z_0)}= \frac{1}{(\omega - z_0)\left(1 -\frac{z - z_0}{\omega - z_0}\right)} \\[4pt] &= \frac{1}{(\omega - z_0)}\sum_{n=0}^\infty \left(\frac{z - z_0}{\omega - z_0}\right)^n,\end{split}$

and the latter converges absolutely and uniformly for $$|z - z_0| < |\omega - z_0|$$.

Now fix $$z_0\in \mathbb C \setminus \gamma^*$$. Then, for all $$z$$ satisfying $$|z - z_0| < \mathrm{dist}(z_0, \gamma^*)$$,

$\begin{split}F(z) &= \int_\gamma \frac{\varphi(\omega)}{(\omega-z)} \, d\omega = \int_\gamma \frac{\varphi(\omega)}{(\omega-z_0)}\sum_{n=0}^\infty\left(\frac{z - z_0}{\omega - z_0}\right)^n \, d\omega\\[4pt] &= \sum_{n=0}^\infty \left(\int_\gamma \frac{\varphi(\omega)}{(\omega-z_0)^{n+1}}\, d\omega\right) (z - z_0)^n = \sum_{n=0}^\infty a_n (z - z_0)^n,\end{split}$

where $$a_n = \int_\gamma \frac{\varphi(\omega)}{(\omega-z_0)^{n+1}}\, d\omega$$.

(Note: interchanging the summation and integration is okay since the sum converges absolutely and uniformly.)

Since we can represent $$F$$ as such a power series about every $$z_0 \in \mathbb C \setminus \gamma^*$$, this proves $$F \in H(\mathbb C \setminus \gamma^*)$$.

By Taylor’s theorem, if $$F(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$$, then $$a_n = \frac{F^{(n)}(z_0)}{n!}$$.

Comparing this with the result above, we see that

$F^{(n)}(z_0) = n! \int_\gamma \frac{\varphi(\omega)}{(\omega - z_0)^{n+1}}\, d\omega,$

holds for every $$z\in \mathbb C \setminus \gamma^*$$. In particular,

$F'(z_0) = \int_\gamma \frac{\varphi(\omega)}{(\omega - z_0)^2}\, d\omega.$

Proof 2. We first recall two easy facts.

Fact 1. $$\varphi$$ is uniformly continuous on the compact set $$\gamma^*$$ and the image $$\varphi[\gamma^*]$$ is compact.

Fact 2. $$F(z)$$ is a Cauchy integral of a continuous function, so $$F \in H(\mathbb C \setminus \gamma^*)$$.

Fix $$z_0\in \mathbb C \setminus \gamma^*$$ and let

$d_0 = \mathrm{dist}(z_0, \gamma^*) = \inf\{|\zeta - z_0| : \zeta \in \gamma^*\}.$

Assume $$0 < |a - z_0| < d_0/2$$. Then

$\begin{split}\frac{F(a) - F(z_0)}{a - z_0} &= \frac{1}{a-z_0} \int_\gamma \frac{\varphi(\omega)}{\omega - a} \, d \omega - \int_\gamma \frac{\varphi(\omega)}{\omega - z_0} \, d \omega\\[4pt] &= \frac{1}{a-z_0} \int_\gamma \frac{\varphi(\omega)[(\omega-z_0) - (\omega-a)]}{(\omega - a)(\omega - z_0)} \, d \omega\\[4pt] &= \int_\gamma \frac{\varphi(\omega)}{(\omega - a)(\omega - z_0)} \, d \omega\\[4pt]\end{split}$

Let $$M = \sup\{|\varphi(\omega)| : \omega \in \gamma^*\}$$ and observe that $$M < \infty$$. Also, for $$\omega \in \gamma$$, we have $$|\omega - a| > d_0/2$$ and $$|\omega - z_0| \geq d_0$$. Therefore,

$\frac{|\varphi(\omega)|}{|\omega - a| |\omega - z_0|} < \frac{M}{\frac{d_0}{2}\cdot d_0}= \frac{2M}{d_0^2}.$

This holds for all $$a \in B(z_0, d_0/2) \setminus \{z_0\}$$. Therefore,

$\begin{split}F'(z) &= \lim_{a \to z_0} \frac{F(a) - F(z_0)}{a - z_0} \\[4pt] &= \lim_{a \to z_0} \int_\gamma \frac{\varphi(\omega)}{(\omega-a)(\omega-z_0)} \, d\omega\\[4pt] &= \int_\gamma \lim_{a \to z_0} \frac{\varphi(\omega)}{(\omega-a)(\omega-z_0)} \, d\omega\\[4pt] &= \int_\gamma \frac{\varphi(\omega)}{(\omega-z_0)^2} \, d\omega.\end{split}$

Solution to Problem 35

1. Theorem. (Casorati-Weierstrass) Let $$f$$ be holomorphic in a region $$G \subset \mathbb C$$ except for at isolated singularity at $$z_0 \in G$$. Then for every $$\omega \in \mathbb C$$ there is a sequence $$\{z_n\} \subset G$$ such that $$z_n \to z_0$$ and $$\lim_{n\to \infty}f(z_n) = \omega$$.

In other words, in a neighborhood $$U$$ of an isolated singularity $$z_0$$, the image set $$f[U \setminus \{z_0\}]$$ is dense in $$\mathbb C$$.

1. We give two solutions.

Solution 1. Consider the expansion of $$(z-3) \sin\bigl(\frac{1}{z+1}\bigr)$$. The factor $$f(z) := \sin\bigl(\frac{1}{z+1}\bigr)$$ has an isolated singularity at $$z_0 = -2$$ and this is the only singularity of $$f$$ in $$|z|\leq R$$.

Now, recall,

$\sin z = \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}.$

Let $$\omega = z+2$$. Then $$z = \omega-2$$, so $$z-3 = \omega-5$$, so

$\begin{split}(z-3) \sin\bigl(\frac{1}{z+1}\bigr) &= (\omega-5)\sin \bigl(\frac{1}{\omega}\bigr)\\ &= (\omega-5)\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \frac{1}{\omega^{2n+1}}\\ &= (\omega-5)\bigl[\frac{1}{\omega} - \frac{1}{3!\omega^3} + \frac{1}{5!\omega^5} - \cdots\bigr] \\ &= \bigl(1 - \frac{1}{3!\omega^2} + \frac{1}{5!\omega^4} - \cdots \bigr) - \bigl(\frac{5}{\omega} - \frac{5}{3!\omega^3} +\frac{5}{5!\omega^5} - \cdots \bigr)\\ &= 1 - \frac{5}{\omega} - \frac{1}{3!\omega^2} + \frac{5}{3!\omega^3} +\frac{1}{5!\omega^4} - \frac{5}{5!\omega^5} -\cdots\\ &= 1 - \frac{5}{z+2} - \frac{1}{3!(z+2)^2} + \frac{5}{3!(z+2)^3} +\frac{1}{5!(z+2)^4} - \cdots.\end{split}$

Thus, $$-2$$ is an essential singularity of $$f$$.

Since $$-2$$ is in the interior of $$\{z : |z| = R\}$$, and since $$-5$$ is the coefficient of $$(z-(-2))^{-1}$$, we have

$\frac{1}{2\pi i} \int_{|z| = R}(z-3)\sin\bigl(\frac{1}{z+2}\bigr) \, dz = -5.$

Solution 2. Let $$\omega = \frac{1}{z + 2}$$. Then $$d\omega = -(z + 2)^{-2}\, dz$$, or $$d\omega = -\omega^{2}\, dz$$; also, $$z = \frac{1}{\omega}-2$$. Therefore, the integral in question is

$\begin{split}I &= \frac{1}{2\pi i} \int_{C} \left(\frac{1}{\omega} - 2 - 3\right) \sin \omega \, \frac{d\omega}{-\omega^2}\\[4pt] &= \frac{1}{2\pi i} \int_{C} \left(\frac{5\omega -1}{\omega^3}\right) \sin \omega \, d\omega\\[4pt] &= \frac{5}{2\pi i} \int_{C} \frac{\sin \omega}{\omega^2} \, d\omega - \frac{1}{2\pi i} \int_{C} \frac{\sin \omega}{\omega^3} \, d\omega\end{split}$

Now, consider the curve $$C := \{\frac{1}{2 + Re^{i\theta}} : 0 \leq \theta \leq 2\pi\}$$, where $$R\geq 4$$. Of course, $$\{2 + Re^{i\theta} : 0 \leq \theta \leq 2\pi\}$$ is the circle of radius $$R$$ centered at 2, so $$\frac{1}{2 + Re^{i\theta}}$$ is the reflection of $$2 + Re^{i\theta}$$ across the unit circle. (Note that $$2 + Re^{i\theta}$$ lies entirely outside the unit circle, since $$R\geq 4$$.)

We see that $$C$$ is a closed circle containing the origin in its interior. Therefore, $$I$$ is computed as

$I = 5\mathrm{Res}\bigl(\frac{\sin \omega}{\omega^2}, 0\bigr) - \mathrm{Res}\bigl( \frac{\sin \omega}{\omega^3}, 0\bigr).$

To compute the residues, we recall the Laurent expansion of a function $$f$$ with a pole of order $$m$$ at $$z = z_0$$. That is,

$f(z) = \sum_{n=-m}^\infty a_n (z - z_0)^n.$

It follows that

$(z-z_0)^m f(z) = \sum_{n=-m}^\infty a_n (z - z_0)^{n+m} = a_{-m} + a_{-m+1}(z-z_0) + a_{-m+2}(z-z_0)^2 + \cdots,$

so $$\frac{d}{dz}(z-z_0)^m f(z) = a_{-m+1} + 2a_{-m+2}(z-z_0) + \cdots$$ and

$\left(\frac{d}{dz}\right)^{m-1}(z-z_0)^m f(z) = (m-1)(m-2)\cdots 2a_{-1} + m(m-1)\cdots 2 a_0(z-z_0) + \cdots.$

Thus,

$a_{-1} = \lim_{z\to z_0} \frac{1}{(m-1)!}\left(\frac{d}{dz}\right)^{m-1}(z-z_0)^m f(z) = \mathrm{Res}(f, z_0).$

It follows that

$\mathrm{Res}(\frac{\sin \omega}{\omega^3},0) = \lim_{\omega\to 0} \frac{1}{2}\left(\frac{d}{d\omega}\right)^2\sin \omega= \lim_{\omega\to 0} \frac{-\sin \omega}{2} = 0.$

A similar calculation yields,

$\mathrm{Res}\bigl(\frac{\sin \omega}{\omega^2},0\bigr) = \lim_{\omega\to 0}\frac{d}{d\omega}\sin \omega= \lim_{\omega\to 0} \cos \omega = 1.$

Plugging these residues into the expression for $$I$$ that we derived earlier, we see that $$I = 5$$.

Solution to Problem 36

We give two proofs.

Proof 1.

Since $$f$$ is entire and non-vanishing in $$\mathbb C$$, there exists an entire function $$g$$ such that $$f(z) = e^{g(z)}$$ for all $$z\in \mathbb C$$.

We first show $$g$$ is a constant belonging to the set $$\{i2\pi n : n \in \mathbb Z\}$$.

By (17)

(18)$0 < |f(z)| = |e^{g(z)}| \leq e^{|z|}.$

Let $$g(z) = u(z) + i v(z) = \mathfrak{Re} g(z) + i \mathfrak{Im} g(z)$$. Then, $$|e^{g(z)}| = e^{u(z)}$$. By (18), this implies $$u(z) \leq |z|$$ for all $$z\in \mathbb C$$.

Now, since $$g$$ is entire, we can write it as a power series expansion about $$z=0$$:

(19)$g(z) = \sum_{n=0}^\infty a_n z^n$

and $$f(z) = e^{g(z)}$$ implies $$f'(z) = g'(z)e^{g(z)} = g'(z) f(z)$$. Therefore, the assumptions $$f(0) = 1$$ and $$f'(0) = 0$$ yield $$g'(0)\cdot 1 = 0$$, so $$g'(0) = 0$$.

By the expansion (19), we have $$g'(z) = a_1 + 2a_2 z + \cdots$$. This and $$g'(0) = 0$$ imply $$a_1 = 0$$.

By (18), $$|e^{g(z)}| \leq e^{|z|}$$, that is,

$|e^{a_0 + a_1z + a_2z^2 + \cdots} | \leq e^{|z|},$

which holds iff

$|e^{a_0}| |e^{a_1z}| |e^{a_2z^2}| \cdots \leq e^{|z|}$

In particular, this holds when $$z = a$$ is a large real number, which is only possible if $$a_2 = a_3 = \cdots = 0$$ (and we already showed $$a_1 = 0$$).

Therefore, $$a_0$$ is the only nonzero coefficient in the expansion of $$g(z)$$. In other words, $$g(z)$$ is constant.

We now have $$f(z) = e^{a_0}$$ for some constant $$a_0 \in \mathbb C$$. Therefore, $$f$$ is constant, and, by assumption, $$f(0)=1$$. Therefore, $$f(z) = 1$$ for all $$z\in \mathbb C$$.

Proof 2. 1 By Hadamard’s factorization theorem, an entire function $$f$$ with zeros at $$\{a_n\}\subset \mathbb{C}\setminus \{0\}$$ and $$m$$ zeros at $$z=0$$ has the form

$f(z) = e^{P(z)} z^m \prod_{n=0}^\infty \left(1-\frac{z}{a_n}\right) e^{z/a_n}, :label: had$

where $$P(z)$$ is a polynomial of degree $$\rho$$, the “order of growth,” and $$k\leq \rho < k+1$$. For the function in question, we have $$|f(z)|>0$$ so $$\{a_n\} = \emptyset$$ and $$m=0$$. Also, since $$|f(z)|\leq e^{|z|}$$, the order of growth is $$\rho=1$$, which implies that $$P(z)$$ is a polynomial of degree 1. Therefore, ([eq:had]) takes the simple form $$f(z) = e^{Bz+C}$$, for some constants $$B, C$$. We are given that $$f(0)=1$$ and $$f'(0)=0$$, so $$e^{C} = 1$$, and $$f'(0) = Be^{C} = B = 0$$. It follows that $$f(z) = 1$$.

(An alternative proof of this result appears in Rudin’s Functional Analysis book ([Rud91]) on page 250.)

Solution to Problem 37. (coming soon)

Solution to Problem 38.

Suppose there is a holomorphic bijection of the open punctured unit disk, $$G = \{z \mid 0< z < 1\}$$ onto the annulus, $$A(0; r:R) = \Omega = \{z \mid r < |z < R\}$$.

Then, since $$|f(z)|$$ remains bounded, $$z=0$$ is a removable singularity and $$f \in H(D)$$ where $$f(0) = \lim_{z\to 0}f(z)$$.

Consider the possible values of $$f(0)$$. Since $$f$$ is continuous, $$f(0) \in \{\omega \mid r \leq |\omega| \leq R\}$$.

Also $$f(0) \notin \{\omega \mid |\omega|=r\}\cup \{\omega \mid |\omega|=R\}$$, the boundary of the annulus, since that would violate the open mapping theorem.

(Indeed, $$f$$ maps the open disk $$\{z \mid |z| < 1\}$$ onto the set $$\{w \mid r < |w| < 1\} \cup \{f(0)\}$$, which is not open if $$|f(0)| = r$$ or $$|f(0)| = R$$.)

So, suppose $$f(0) = \omega_0 \in \Omega$$; then, since $$f$$ maps $$G$$ onto $$\Omega$$, there exists $$z_0 \in \Omega \setminus \{0\}$$ such that $$f(z_0) = \omega_0 = f(0)$$.

Let $$U, V$$ be disjoint open subsets of the unit disk such that $$0\in U$$ and $$z_0\in V$$. Then (by the open mapping theorem) $$f(U) \cap f(V)$$ is a nonempty open set.

Therefore, there exists $$\omega_1 \in f(U) \cap f(V)$$ such that $$\omega_1 \neq \omega_0$$. Thus, there exists $$z_1 \in U$$ such that $$f(z_1) = \omega_1$$ and, of course, $$z_1 \notin \{0, z_0\}$$, since $$f(z_1) \neq \omega_0$$.

Similarly, $$\omega_1 \in f(U) \cap f(V)$$ implies there exists $$z_2 \in V$$ such that $$f(z_2) = \omega_1$$ and $$z_2 \notin \{0, z_0, z_1\}$$.

But then we have $$f(z_1) = \omega_1 = f(z_2)$$ and $$z_1 \neq z_2$$, contradicting the assumption that $$f$$ is a bijection (in particular, one-to-one).

Solution to Problem 39.

1. See Montel’s theorem in the Appendix.

2. By the root test, $$\sum_{n=0}^\infty M_n < \infty$$. Therefore, for every $$0\leq r < 1$$, we have,

$|f(z)| \leq \sum_{n=0}^\infty |a_n| |z|^n \leq \sum_{n=0}^\infty |a_n| \leq \sum_{n=0}^\infty M_n < \infty,$

for all $$|z| < r$$ and all $$f \in \mathcal F$$.

Since this is stronger than the sufficient condition in Montel’s theorem (i.e., local boundedness), $$\mathcal F$$ is a normal family.

Solution to Problem 40. (coming soon)

Solution to Problem 41. (coming soon)

Footnotes

1

I was fortunate to have worked on this exam after having just read a beautiful treatment of the Hadamard factorization theorem in Stein and Sharkachi’s book [SS03]. If you need convincing that this theorem is worth studying, take a look at how easily it dispenses with this and other, otherwise challenging exam problems. Stein and Sharkachi seem to have set things up just right, so that the theorem is very easy to apply.     Real Analysis Exams