# 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.

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

Suppose that we do

*not*assume that \(u\) and \(v\) are \(C^1\), but merely that \(u\) and \(v\) are continuous in \(G\) and

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

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 \(T\).

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.

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 \(z_0\).

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

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 \(2\pi i\).

From the theory of Laurent expansions, it is known that there are constants \(a_n\) such that, for \(1<|z|<4\),

Find \(a_{-10}\) and \(a_{10}\) by the method of your choice.

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

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

(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

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

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 \(1\leq R \leq |z| <\infty\),

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

**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

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.

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

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.

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

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, \(2\pi R\).

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.

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