.. 2001 Nov 26 .. =========== 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, :math:`z` denotes a complex variable, and :math:`\mathbb{C}` denotes the complex plane. .. index:: Cauchy-Riemann equations, Cauchy's theorem, Green's theorem .. _2001Nov_1: .. proof:prob:: (a) Suppose that :math:`f(z) = f(x+iy) = u(x,y) + i v(x,y)` where :math:`u` and :math:`v` are :math:`C^1` functions defined on a neighborhood of the closure of a bounded region :math:`G\subset \mathbb{C}` with boundary which is parametrized by a properly oriented, piecewise :math:`C^1` curve :math:`\gamma`. If :math:`u` and :math:`v` obey the Cauchy-Riemann equations, show that Cauchy's theorem :math:`\int_\gamma f(z) \, dz = 0` follows from Green's theorem, namely .. math:: \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 $Q$.} :label: 200111 (b) Suppose that we do *not* assume that :math:`u` and :math:`v` are :math:`C^1`, but merely that :math:`u` and :math:`v` are continuous in :math:`G` and .. math:: f'(z_0) = \lim_{z\rightarrow z_0} \frac{f(z) - f(z_0)}{z-z_0} exists at some (possibly only one!) point :math:`z_0 \in G`. Show that given any :math:`\epsilon >0`, we can find a triangular region :math:`\Delta` containing :math:`z_0`, such that if :math:`T` is the boundary curve of :math:`\Delta`, then .. math:: \left|\int_T f(z)\, dz\right| = \frac{1}{2}\epsilon L^2, where :math:`L` is the length of the perimeter of :math:`\Delta`. **Hint for** (b) Note that part (a) yields :math:`\int_T (az+b) \, dz =0` for :math:`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 :math:`\left|\int_T g(z)\, dz\right| \leq L \cdot \sup\{|g(z)|:z\in T\}` for :math:`g` continuous on :math:`T`. .. index:: Liouville's theorem, argument principle, maximum principle, Rouché's theorem, open mapping theorem, fundamental theorem of algebra .. _2001Nov_2: .. proof:prob:: 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. .. index:: Casorati-Weierstrass theorem .. _2001Nov_3: .. proof:prob:: (a) 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 :math:`g` is analytic and bounded in the neighborhood of a point :math:`z_0`, then :math:`g` has a removable singularity at :math:`z_0`. (b) Verify the Casorati-Weierstrass theorem directly for a specific analytic function of your choice, with a suitable singularity. .. index:: residue theorem, Laurent expansion .. _2001Nov_4: .. proof:prob:: (a) Define :math:`\gamma : [0,2\pi] \rightarrow \mathbb{C}` by :math:`\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 :math:`f` such that :math:`\int_\gamma f(z) \, dz = 1`. Be careful with minus signs and factors of :math:`2\pi i`. (b) From the theory of Laurent expansions, it is known that there are constants :math:`a_n` such that, for :math:`1<|z|<4`, .. math:: \frac{1}{z^2 - 5z + 4} = \sum_{n=-\infty}^\infty a_n z^n. Find :math:`a_{-10}` and :math:`a_{10}` by the method of your choice. .. index:: Arzela-Ascoli theorem, Montel's theorem, Cauchy's formula .. _2001Nov26_5: .. _2001Nov_5: .. proof:prob:: (a) Suppose that :math:`f` is analytic on a region :math:`G\subset \mathbb{C}` and :math:`\{z\in \mathbb{C}: |z-a|\leq R\} \subset G`. Show that if :math:`|f(z)| \leq M` for all :math:`z` with :math:`|z-a|=R`, then for any :math:`w_1, w_2\in \{z\in \mathbb{C}: |z-a|\leq \frac{1}{2}R\}`, we have .. math:: |f(w_1) - f(w_2)| \leq \frac{4M}{R} |w_1 - w_2| (b) 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 :math:`F` of analytic functions on a region :math:`G`. -------------------------------------- Solutions ---------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2001Nov_1>` (a) Let :math:`P=u` and :math:`Q= -v` in :eq:`200111`. Then, by the Cauchy-Riemann equations, [1]_ .. math:: \int_\gamma u(x,y)\, dx - v(x,y) \, dy= \int_G (v_x + u_y) \, dx\, dy = 0. :label: 200112 Similarly, if :math:`P=v` and :math:`Q=u` in Green’s theorem, then the Cauchy-Riemann equations imply .. math:: \int_\gamma v(x,y)\, dx + u(x,y) \, dy= \int_G (u_x - v_y) \, dx\, dy = 0. :label: 200113 Next, note that .. math:: 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 :eq:`200112` and :eq:`200113`, .. math:: \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.   (b) Suppose :math:`u` and :math:`v` are continuous and :math:`f'(z)` exists at the point :math:`z_0\in G`. Then, for any :math:`\epsilon>0` there is a :math:`\delta>0` such that :math:`B(z_0,\delta)\subseteq G`, and .. math:: \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 :math:`\Delta \subset B(z_0,\delta)` with :math:`z_0\in \Delta`, and let :math:`T` be the boundary. Define .. math:: R(z) = f(z) - [f(z_0) + f'(z_0) (z-z_0)]. Then, by Cauchy's theorem (part (a)), :math:`\int_T [f(z_0) + f'(z_0) (z-z_0)] \, dz = 0`, whence :math:`\int_T R(z)\, dz = \int_T f(z)\, dz`. Finally, note that .. math:: \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, .. math:: \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. where :math:`L` denotes the length of the perimeter of :math:`\Delta` (i.e.,the length of :math:`T`), and :math:`r` denotes the length of one side of :math:`T`, which must, of course, be greater than :math:`|z-z_0|` for all :math:`z\in T`. Also, the length of one side of :math:`\Delta` is surely less than half the length of the perimeter (i.e.,\ :math:`r` In the two proofs below, we begin by supposing :math:`p(z)` is not constant and thus has the form :math:`p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n` with :math:`a_n \neq 0` for some :math:`n\geq 1`. Both proofs also rely on the following observation: If :math:`\{a_j\}_{j=0}^n \subset \mathbb{C}` with :math:`a_n\neq 0`, then for all :math:`1\leq R \leq |z| <\infty`, .. math:: \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 ` (a) If :math:`f` is a holomorphic function in a region :math:`G\in \mathbb{C}` except for an essential singularity at the point :math:`z=z_0`, then for any :math:`w\in \mathbb{C}` there is a sequence :math:`\{z_n\}\subset G` approaching :math:`z_0` such that :math:`f(z_n)\rightarrow w` as :math:`n\rightarrow \infty`. **Proof.** Fix :math:`w_0\in \mathbb{C}` and suppose there is no sequence :math:`\{z_n\}\subset G` approaching :math:`z_0` such that :math:`f(z_n)\rightarrow w_0` as :math:`n\rightarrow \infty`. Then there is a punctured disk :math:`\bar{D_0} \triangleq B(z_0,\epsilon)\setminus \{z_0\} \subset G` such that :math:`|f(z)-w_0|>\delta >0` for all :math:`z\in \bar{D_0}`. Define :math:`g(z) = 1/(f(z) - w_0)` on :math:`D_0`. Then .. math:: \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. Thus, by lemma [lem:removable-singularity] (Nov. ’06, prob. 1), :math:`z_0` is a removable singularity of :math:`g(z)`. Therefore, :math:`g(z) \in H(B(z_0, \epsilon))`. In particular, :math:`g` is continuous and non-zero at :math:`z=z_0`, so it is non-zero in a neighborhood :math:`B(z_0,\epsilon_0)` of :math:`z_0`. Therefore, :math:`f(z)-w_0 = 1/g(z)` is holomorphic in :math:`B(z_0,\epsilon_0)`, which implies that the singularity of :math:`f(z)` at :math:`z=z_0` is removable. This contradiction proves the theorem.   (b) Consider :math:`f(z) = e^z`. This function has an essential singularity at :math:`\infty`, and, for every horizontal strip, .. math:: S_\alpha = \{x+iy: x\in \mathbb R, \, \alpha \leq y < \alpha + 2\pi\}, of width :math:`2\pi`, :math:`f(z)` maps :math:`S_\alpha` onto :math:`\mathbb{C}\setminus \{0\}`. (In particular, :math:`f(z)` comes arbitrarily close to every :math:`w\in \mathbb{C}`.) Now let :math:`\mathcal{N}_R = \{z\in \mathbb{C}: |z|>R\}` be any neighborhood of :math:`\infty`. There is clearly a strip :math:`S_\alpha` contained in :math:`\mathcal{N}_R` (e.g.,with :math:`\alpha = R+1`). Therefore, :math:`f(z)=e^z` maps points in :math:`\mathcal{N}_R` to points arbitrarily close (in fact equal when :math:`w\neq 0`) to all points :math:`w\in \mathbb{C}`. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2001Nov_4>` (a) Let :math:`G` be the region whose boundary is the curve :math:`\gamma`, and suppose :math:`f(z) \in H(\mathbb{C})` except for isolated singularities at the points :math:`\{z_1, \ldots, z_n\}\subset G`. By the residue theorem, .. math:: \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 :math:`f(z)\in H(\mathbb{C})` with exactly two isolated singularities in :math:`G` (e.g.,at :math:`z_1=i` and :math:`z_2=-i`), and such that :math:`\mbox{Res}(f,z_j)=\frac{-i}{4\pi}`, then .. math:: \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, .. math:: 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.   (b) Expand the function in partial fractions: .. math:: \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 .. math:: \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 :math:`|z|<4`, while .. math:: \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 :math:`|z|>1`. Therefore, .. math:: \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 $1<|z|<4$.} .. math:: \therefore \quad a_{-10} = -\frac{1}{3} \qquad \text{and} \qquad a_{10} = -\frac{1}{12}\left(\frac{1}{4}\right)^{10}. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2001Nov_5>` (a) By :ref:`Cauchy's formula `, if :math:`w` is any point in the disk :math:`|w-a|< R`, then .. math:: f(w) = \frac{1}{2\pi i} \int_{|\zeta - a| = R} \frac{f(\zeta)}{\zeta-w}\, d\zeta. In particular, if :math:`w_1, w_2` are any two points inside the “half-disk” :math:`|w-a|< R/2`, then .. math:: 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. Now, for all :math:`\zeta` on the outer radius in figure [fig:concentriccircles], it is clear that :math:`|\zeta - w_1|>R/2` and :math:`|\zeta - w_2|>R/2`. Therefore, .. math:: |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|, where :math:`\gamma` denotes the positively oriented circle :math:`\{\zeta: |\zeta - a| = R\}`, and :math:`\ell(\gamma)` denotes its length, :math:`2\pi R`. (b) [4]_ The statements of the :ref:`Arzela-Ascoli theorem ` and :ref:`Montel's theorem ` are given in the :ref:`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 :math:`\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 :cite:`Conway:1978`, page 153. Let :math:`S = \mathbb{C}` in the Arzela-Ascoli theorem. In that case, :math:`K\subset \mathbb{C}` is compact if and only if :math:`K` is closed and bounded. Therefore, if :math:`\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 :math:`a\in G` there is a neighborhood :math:`B(a,r)` in which :math:`\mathcal{F}` is equicontinuous with equicontinuity constant :math:`\delta`. (Why is this sufficient?) [5]_ [6]_ So, fix :math:`a\in G` and :math:`\epsilon >0`, and let :math:`\bar{B}(a,R)\subset G`. Then, by local boundedness, there is an :math:`M>0` such that :math:`|f(z)|\leq M` for all :math:`z \in \bar{B}(a,R)` and all :math:`f\in \mathcal{F}`. Therefore, by part (a), .. math:: |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 :math:`\delta = \frac{R}{4M}\epsilon` and :math:`r = R/2`, then :math:`|f(w_1)-f(w_2)| < \epsilon` whenever :math:`w_1, w_2 \in B(a,r)` and :math:`|w_1-w_2|< \delta`. Therefore, :math:`\mathcal{F}` is equicontinuous in :math:`B(a,r)`. We have thus shown that local boundedness implies conditions (i) and (ii) of the Arzela-Ascoli theorem and thereby implies normality. -------------------------- .. rubric:: Footnotes .. [1] These are :math:`u_x = v_y` and :math:`u_y = -v_x`. .. [2] Note, we add 1 here just to be sure :math:`R` is safely over 1. .. [3] Conway :cite:`Conway:1978`, page 77, presents a similar, but more elegant proof. .. [4] The best treatment of normal families and the Arzela-Ascoli theorem is Ahlfors :cite:`Ahlfors:1968`. .. [5] Answer: If, instead of a single point :math:`a\in G`, we are given a compact set :math:`K\subset G`, then there is a finite cover :math:`\{B(a_j,r_j):j=1,\ldots, n\}` by such neighborhoods with equicontinuity constants :math:`\delta_1, \ldots, \delta_n`. Then, :math:`\delta = \min_j\delta_j`, is a single equicontinuity constant that works for all of :math:`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 :cite:`Ahlfors:1968` and Rudin :cite:`Rudin:1987`, and the “pointwise” equicontinuity discussed in topology books like the one by Munkres :cite:`Munkres:2000`. 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. ------------------------------------- .. blank