# Examination 5¶

## 2001 Nov¶

Instructions. Make a substantial effort on all parts of the following problems. If you cannot completely answer Part (a) of a problem, it is still possible to do Part (b). Partial credit is given for partial progress. Include as many details as time permits. Throughout the exam, $$z$$ denotes a complex variable, and $$\mathbb{C}$$ denotes the complex plane.

Problem 29
1. Suppose that $$f(z) = f(x+iy) = u(x,y) + i v(x,y)$$ where $$u$$ and $$v$$ are $$C^1$$ functions defined on a neighborhood of the closure of a bounded region $$G\subset \mathbb{C}$$ with boundary which is parametrized by a properly oriented, piecewise $$C^1$$ curve $$\gamma$$. If $$u$$ and $$v$$ obey the Cauchy-Riemann equations, show that Cauchy’s theorem $$\int_\gamma f(z) \, dz = 0$$ follows from Green’s theorem, namely

(11)$\int_\gamma P\,dx + Q\,dy = \int_G \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \, dx\, dy \;\text{ for C^1 functions P and Q.}$
1. Suppose that we do not assume that $$u$$ and $$v$$ are $$C^1$$, but merely that $$u$$ and $$v$$ are continuous in $$G$$ and

$f'(z_0) = \lim_{z\rightarrow z_0} \frac{f(z) - f(z_0)}{z-z_0}$

exists at some (possibly only one!) point $$z_0 \in G$$. Show that given any $$\epsilon >0$$, we can find a triangular region $$\Delta$$ containing $$z_0$$, such that if $$T$$ is the boundary curve of $$\Delta$$, then

$\left|\int_T f(z)\, dz\right| = \frac{1}{2}\epsilon L^2,$

where $$L$$ is the length of the perimeter of $$\Delta$$.

Hint for (b) Note that part (a) yields $$\int_T (az+b) \, dz =0$$ for $$a, b \in \mathbb{C}$$, which you can use here in (b), even if you could not do Part (a). You may also use the fact that $$\left|\int_T g(z)\, dz\right| \leq L \cdot \sup\{|g(z)|:z\in T\}$$ for $$g$$ continuous on $$T$$.

Problem 30

Give two quite different proofs of the fundamental theorem of algebra, that if a polynomial with complex coefficients has no complex zero, then it is constant. You may use independent, well-known theorems and principles such as Liouville’s theorem, the argument principle, the maximum principle, Rouché’s theorem, and/or the open mapping theorem.

Problem 31
1. State and prove the Casorati-Weierstrass theorem concerning the image of any punctured disk about a certain type of isolated singularity of an analytic function. You may use the fact that if a function $$g$$ is analytic and bounded in the neighborhood of a point $$z_0$$, then $$g$$ has a removable singularity at $$z_0$$.

2. Verify the Casorati-Weierstrass theorem directly for a specific analytic function of your choice, with a suitable singularity.

Problem 32
1. Define $$\gamma : [0,2\pi] \rightarrow \mathbb{C}$$ by $$\gamma(t) = \sin (2t) + 2i \sin (t)$$. This is a parametrization of a “figure 8” curve, traced out in a regular fashion. Find a meromorphic function $$f$$ such that $$\int_\gamma f(z) \, dz = 1$$. Be careful with minus signs and factors of $$2\pi i$$.

2. From the theory of Laurent expansions, it is known that there are constants $$a_n$$ such that, for $$1<|z|<4$$,

$\frac{1}{z^2 - 5z + 4} = \sum_{n=-\infty}^\infty a_n z^n.$

Find $$a_{-10}$$ and $$a_{10}$$ by the method of your choice.

Problem 33
1. Suppose that $$f$$ is analytic on a region $$G\subset \mathbb{C}$$ and $$\{z\in \mathbb{C}: |z-a|\leq R\} \subset G$$. Show that if $$|f(z)| \leq M$$ for all $$z$$ with $$|z-a|=R$$, then for any $$w_1, w_2\in \{z\in \mathbb{C}: |z-a|\leq \frac{1}{2}R\}$$, we have

$|f(w_1) - f(w_2)| \leq \frac{4M}{R} |w_1 - w_2|$
1. Explain how Part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting the normality of any locally bounded family $$F$$ of analytic functions on a region $$G$$.

## Solutions¶

Solution to Problem 29

1. Let $$P=u$$ and $$Q= -v$$ in (11). Then, by the Cauchy-Riemann equations, 1

(12)$\int_\gamma u(x,y)\, dx - v(x,y) \, dy= \int_G (v_x + u_y) \, dx\, dy = 0.$

Similarly, if $$P=v$$ and $$Q=u$$ in Green’s theorem, then the Cauchy-Riemann equations imply

(13)$\int_\gamma v(x,y)\, dx + u(x,y) \, dy= \int_G (u_x - v_y) \, dx\, dy = 0.$

Next, note that

$f(z)\, dz = [u(x,y)+iv(x,y)]\, d(x+iy) = u(x,y)\, dx - v(x,y)\, dy +i [v(x,y)\, dx + u(x,y)\, dy].$

Therefore, by (12) and (13),

$\int_\gamma f(z)\, dz = \int_\gamma u(x,y)\, dx - v(x,y)\, dy +i \int_\gamma v(x,y)\, dx + u(x,y)\, dy = 0.$
1. Suppose $$u$$ and $$v$$ are continuous and $$f'(z)$$ exists at the point $$z_0\in G$$. Then, for any $$\epsilon>0$$ there is a $$\delta>0$$ such that $$B(z_0,\delta)\subseteq G$$, and

$\left| f'(z_0) - \frac{f(z) - f(z_0)}{z-z_0}\right| < \epsilon, \quad \text{ for all |z- z_0|< \delta.}$

Pick a triangular region $$\Delta \subset B(z_0,\delta)$$ with $$z_0\in \Delta$$, and let $$T$$ be the boundary. Define

$R(z) = f(z) - [f(z_0) + f'(z_0) (z-z_0)].$

Then, by Cauchy’s theorem (part (a)), $$\int_T [f(z_0) + f'(z_0) (z-z_0)] \, dz = 0$$, whence $$\int_T R(z)\, dz = \int_T f(z)\, dz$$. Finally, note that

$\left|\frac{R(z)}{z-z_0}\right| = \left| f'(z_0) - \frac{f(z) - f(z_0)}{z-z_0}\right| < \epsilon, \quad \text{ for all |z- z_0|< \delta.}$

Therefore,

$\begin{split}\left|\int_T f(z)\, dz\right| &= \left|\int_T R(z)\, dz \right| \leq \int_T |R(z)|\, |dz| \\[4pt] &= \int_T \left|\frac{R(z)}{z-z_0}\right| |z-z_0|\, |dz| \\[4pt] &\leq \epsilon \int_T |z-z_0|\, |dz| \leq \epsilon r L.\end{split}$

where $$L$$ denotes the length of the perimeter of $$\Delta$$ (i.e.,the length of $$T$$), and $$r$$ denotes the length of one side of $$T$$, which must, of course, be greater than $$|z-z_0|$$ for all $$z\in T$$. Also, the length of one side of $$\Delta$$ is surely less than half the length of the perimeter (i.e.,$$r<L/2$$). Therefore,

$\left|\int_T f(z)\, dz\right| \leq \frac{1}{2}\epsilon L^2.$

Solution to Problem 30

In the two proofs below, we begin by supposing $$p(z)$$ is not constant and thus has the form $$p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$. Both proofs also rely on the following observation: If $$\{a_j\}_{j=0}^n \subset \mathbb{C}$$ with $$a_n\neq 0$$, then for all $$1\leq R \leq |z| <\infty$$,

$\begin{split}\left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| &\leq \frac{|a_0|}{|a_n|} |z|^{-n} + \cdots +\frac{|a_{n-1}|}{|a_n|} |z|^{-1}\\[4pt] &\leq n \max_{0\leq j <n} \frac{|a_j|}{|a_n|}|z|^{-1} \\[4pt] &\leq n \max_{0\leq j <n} \frac{|a_j|}{|a_n|}R^{-1}.\end{split}$

In particular, if we choose 2 $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$, then

(14)$\left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| \leq 1/2,\quad \text{ for all |z| \geq R.}$

Proof 1.

Assume $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and let $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$, as above. We claim that

(15)$|p(z) - a_n z^n| < |a_n z^n|, \quad \text{ for all } |z|=R.$

To see this, check that

$\frac{|p(z) - a_n z^n|}{|a_n z^n|} = \left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| < 1, \quad \text{ for all |z|=R.}$

In fact, (14) implies that the sum is no greater than 1/2, for all $$|z|\geq R$$, which is more than we need.

Now (15) and Rouché’s theorem imply that the function $$g(z) = a_nz^n$$ has the same number of zeros in $$|z|<R$$ as does the function $$p(z)$$. Clearly $$z=0$$ is a zero of $$g(z)$$ (of multiplicity $$n$$). Therefore, $$p(z)$$ has a zero in $$|z|<R$$.

Proof 2.

3 Assume $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and consider

(16)$|p(z)| = |a_n z^n| \left|\frac{a_0}{a_n}z^{-n} + \cdots + \frac{a_{n-1}}{a_n} z^{-1} + 1\right|\geq |a_n||z|^n \left| 1 - |\sum_{j=0}^{n-1}\frac{a_j}{a_n}z^{-n+j}|\,\right|.$

If we choose $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$ as above, then for all $$|z|\geq R$$,

$0\leq \left|\sum_{j=0}^{n-1}\frac{a_j}{a_n}z^{-n+j}\right| = \left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| \leq 1/2,$

and (16) yields $$|p(z)| \geq |a_n||z|^n/2$$, for all $$|z|\geq R$$. Therefore, the function $$f(z) \triangleq 1/p(z)$$ satisfies

$|f(z)| = \frac{1}{|p(z)|} \leq \frac{2}{|a_n||z|^n}, \quad \text{ for all |z|\geq R.}$

Now suppose $$p(z)$$ has no complex zero. Then $$f(z) \in H(\mathbb{C})$$. In particular, $$f(z)$$ is continuous, hence bounded on the compact set $$|z|\leq R$$. Therefore $$f(z)$$ is a bounded entire function, so, by Liouville’s theorem, it must be constant, but then $$p(z)$$ must be constant. This contradicts our initial assumption and proves that $$p(z)$$ must have a complex zero.

In fact, we have proved a bit more: If $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and $$R$$ is either 1 or $$R = 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$ (whichever is greater), then $$p(z)$$ vanishes for some $$|z|< R$$, while for all $$|z|\geq R$$,:math:|p(z)| is bounded from below by $$|a_n||z|^n/2$$. Thus all the zeros of $$p(z)$$ are contained in the disk $$|z|< R$$.

Solution to Problem 31

1. If $$f$$ is a holomorphic function in a region $$G\in \mathbb{C}$$ except for an essential singularity at the point $$z=z_0$$, then for any $$w\in \mathbb{C}$$ there is a sequence $$\{z_n\}\subset G$$ approaching $$z_0$$ such that $$f(z_n)\rightarrow w$$ as $$n\rightarrow \infty$$.

Proof. Fix $$w_0\in \mathbb{C}$$ and suppose there is no sequence $$\{z_n\}\subset G$$ approaching $$z_0$$ such that $$f(z_n)\rightarrow w_0$$ as $$n\rightarrow \infty$$. Then there is a punctured disk $$\bar{D_0} \triangleq B(z_0,\epsilon)\setminus \{z_0\} \subset G$$ such that $$|f(z)-w_0|>\delta >0$$ for all $$z\in \bar{D_0}$$.

Define $$g(z) = 1/(f(z) - w_0)$$ on $$D_0$$. Then

$\begin{split}\limsup_{\substack{z\rightarrow z_0\\ z\in D_0}} |g(z)| = \limsup_{\substack{z\rightarrow z_0\\ z\in D_0}} \frac{1}{|f(z) - w_0|}\leq \frac{1}{\delta} < \infty.\end{split}$

Thus, by lemma [lem:removable-singularity] (Nov. ’06, prob. 1), $$z_0$$ is a removable singularity of $$g(z)$$. Therefore, $$g(z) \in H(B(z_0, \epsilon))$$.

In particular, $$g$$ is continuous and non-zero at $$z=z_0$$, so it is non-zero in a neighborhood $$B(z_0,\epsilon_0)$$ of $$z_0$$. Therefore, $$f(z)-w_0 = 1/g(z)$$ is holomorphic in $$B(z_0,\epsilon_0)$$, which implies that the singularity of $$f(z)$$ at $$z=z_0$$ is removable. This contradiction proves the theorem.

1. Consider $$f(z) = e^z$$. This function has an essential singularity at $$\infty$$, and, for every horizontal strip,

$S_\alpha = \{x+iy: x\in \mathbb R, \, \alpha \leq y < \alpha + 2\pi\},$

of width $$2\pi$$, $$f(z)$$ maps $$S_\alpha$$ onto $$\mathbb{C}\setminus \{0\}$$. (In particular, $$f(z)$$ comes arbitrarily close to every $$w\in \mathbb{C}$$.)

Now let $$\mathcal{N}_R = \{z\in \mathbb{C}: |z|>R\}$$ be any neighborhood of $$\infty$$. There is clearly a strip $$S_\alpha$$ contained in $$\mathcal{N}_R$$ (e.g.,with $$\alpha = R+1$$).

Therefore, $$f(z)=e^z$$ maps points in $$\mathcal{N}_R$$ to points arbitrarily close (in fact equal when $$w\neq 0$$) to all points $$w\in \mathbb{C}$$.

Solution to Problem 32

1. Let $$G$$ be the region whose boundary is the curve $$\gamma$$, and suppose $$f(z) \in H(\mathbb{C})$$ except for isolated singularities at the points $$\{z_1, \ldots, z_n\}\subset G$$. By the residue theorem,

$\int_\gamma f(z) \, dz = 2\pi i \sum_{j=1}^n \mbox{Res}(f,z_j).$

Therefore, if we were to find a function $$f(z)\in H(\mathbb{C})$$ with exactly two isolated singularities in $$G$$ (e.g.,at $$z_1=i$$ and $$z_2=-i$$), and such that $$\mbox{Res}(f,z_j)=\frac{-i}{4\pi}$$, then

$\int_\gamma f(z) \, dz = 2\pi i \sum_j \mbox{Res}(f,z_j) = 2\pi i \left(\frac{-i}{4\pi}+\frac{-i}{4\pi}\right) = 1,$

and the problem would be solved. Clearly,

$f(z) = \frac{-i}{4\pi}\left(\frac{1}{z-i} - \frac{1}{z+i}\right) = \frac{1}{2\pi i} \frac{z}{z^2+1}$

is such a function.

2. Expand the function in partial fractions:

$\frac{1}{z^2 - 5z + 4} = \frac{1}{(z-4)(z-1)} = \frac{1/3}{z-4}- \frac{1/3}{z-1}.$

Then, note that

$\frac{1/3}{z-4} = \frac{1}{3}\frac{-1}{4(1-z/4)} = -\frac{1}{12} \sum_{n=0}^\infty \left(\frac{z}{4}\right)^n$

converges for $$|z|<4$$, while

$\frac{1/3}{z-1} = -\frac{1}{3}\frac{1}{z(1-1/z)} = -\frac{1}{3z} \sum_{n=0}^\infty z^{-n}$

converges for $$|z|>1$$. Therefore,

$\frac{1}{z^2 - 5z + 4} =-\frac{1}{3} \sum_{n=-\infty}^{-1} z^{n}-\frac{1}{12} \sum_{n=0}^\infty \left(\frac{1}{4}\right)^n z^n, \quad \text{ for 1<|z|<4.}$
$\therefore \quad a_{-10} = -\frac{1}{3} \qquad \text{and} \qquad a_{10} = -\frac{1}{12}\left(\frac{1}{4}\right)^{10}.$

Solution to Problem 33

1. By Cauchy’s formula, if $$w$$ is any point in the disk $$|w-a|< R$$, then

$f(w) = \frac{1}{2\pi i} \int_{|\zeta - a| = R} \frac{f(\zeta)}{\zeta-w}\, d\zeta.$

In particular, if $$w_1, w_2$$ are any two points inside the “half-disk” $$|w-a|< R/2$$, then

$\begin{split}f(w_1) - f(w_2) &= \frac{1}{2\pi i} \int_{|\zeta - a| = R} \left[\frac{f(\zeta)}{\zeta-w_1} - \frac{f(\zeta)}{\zeta-w_2}\right]\, d\zeta\\[4pt] &= \frac{w_1 - w_2}{2\pi i}\int_{|\zeta - a| = R} \frac{f(\zeta)}{(\zeta-w_1)(\zeta-w_2)}\, d\zeta.\end{split}$

Now, for all $$\zeta$$ on the outer radius in figure [fig:concentriccircles], it is clear that $$|\zeta - w_1|>R/2$$ and $$|\zeta - w_2|>R/2$$. Therefore,

$\begin{split}|f(w_1) - f(w_2)| &\leq \frac{|w_1 - w_2|}{2\pi}\int_{|\zeta - a| = R} \frac{|f(\zeta)|}{(R/2)^2}\, |d\zeta|\\[4pt] &\leq \frac{|w_1 - w_2|}{2\pi}\, \frac{\sup_\gamma |f(\zeta)|}{R^2/4}\,\ell(\gamma) \\[4pt] &\leq \frac{4M}{R}|w_1 - w_2|,\end{split}$

where $$\gamma$$ denotes the positively oriented circle $$\{\zeta: |\zeta - a| = R\}$$, and $$\ell(\gamma)$$ denotes its length, $$2\pi R$$.

2. 4 The statements of the Arzela-Ascoli theorem and Montel’s theorem are given in the Appendix.

We must explain how part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting the normality of any locally bounded family $$\mathcal{F}\subset H(G)$$.

Because of the way the problem is stated, it is probably enough to prove just one direction of Montel’s theorem; i.e., local boundedness implies normality. For a proof of the other direction, see Conway [Con78], page 153.

Let $$S = \mathbb{C}$$ in the Arzela-Ascoli theorem. In that case, $$K\subset \mathbb{C}$$ is compact if and only if $$K$$ is closed and bounded. Therefore, if $$\mathcal{F}$$ is locally bounded, condition (ii) of the theorem is clearly satisfied. To check that local boundedness also implies condition (i), we use part (a).

It suffices to prove that for any $$a\in G$$ there is a neighborhood $$B(a,r)$$ in which $$\mathcal{F}$$ is equicontinuous with equicontinuity constant $$\delta$$. (Why is this sufficient?) 5 6

So, fix $$a\in G$$ and $$\epsilon >0$$, and let $$\bar{B}(a,R)\subset G$$. Then, by local boundedness, there is an $$M>0$$ such that $$|f(z)|\leq M$$ for all $$z \in \bar{B}(a,R)$$ and all $$f\in \mathcal{F}$$. Therefore, by part (a),

$|f(w_1)-f(w_2)| \leq \frac{4M}{R}|w_1-w_2|, \quad \text{ for all w_1, w_2 \in \{|w-a|\leq R/2\}.}$

If $$\delta = \frac{R}{4M}\epsilon$$ and $$r = R/2$$, then $$|f(w_1)-f(w_2)| < \epsilon$$ whenever $$w_1, w_2 \in B(a,r)$$ and $$|w_1-w_2|< \delta$$. Therefore, $$\mathcal{F}$$ is equicontinuous in $$B(a,r)$$. We have thus shown that local boundedness implies conditions (i) and (ii) of the Arzela-Ascoli theorem and thereby implies normality.

Footnotes

1

These are $$u_x = v_y$$ and $$u_y = -v_x$$.

2

Note, we add 1 here just to be sure $$R$$ is safely over 1.

3

Conway [Con78], page 77, presents a similar, but more elegant proof.

4

The best treatment of normal families and the Arzela-Ascoli theorem is Ahlfors [Ahl78].

5

Answer: If, instead of a single point $$a\in G$$, we are given a compact set $$K\subset G$$, then there is a finite cover $$\{B(a_j,r_j):j=1,\ldots, n\}$$ by such neighborhoods with equicontinuity constants $$\delta_1, \ldots, \delta_n$$. Then, $$\delta = \min_j\delta_j$$, is a single equicontinuity constant that works for all of $$K$$.

6

The careful reader might note the distinction between this type of “uniform” equicontinuity, which is taken for granted in complex analysis texts, e.g.,Ahlfors [Ahl78] and Rudin [Rud87], and the “pointwise” equicontinuity discussed in topology books like the one by Munkres [Mun00]. To make peace with this apparent discrepancy, check that the two notions coincide when the set on which a family of functions is declared equicontinuous is compact.

Real Analysis Exams

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