.. 2007 Apr 16 .. =========== Examination 8 ============= 2007 Apr --------- **Notation.** :math:`\mathbb{C}` is the set of complex numbers, :math:`D= \{z\in \mathbb{C}: |z|<1\}`, and, for any open set :math:`G\subset \mathbb{C}`, :math:`H(G)` is the set of holomorphic functions on :math:`G`. .. index:: Laurent series .. _2007Apr_1: .. proof:prob:: Give the Laurent series expansion of :math:`\frac{1}{z(z-1)}` in the region :math:`A \equiv \{z\in \mathbb{C}: 2< |z+2| < 3\}`. .. _2007Apr_2: .. proof:prob:: (a) Prove: Suppose that for all :math:`z \in D` and all :math:`n\in \mathbb{N}` we have that :math:`f_n` is holomorphic in :math:`D` and :math:`|f_n(z)|<1`. Also suppose that :math:`\lim_{n\rightarrow \infty} \mathfrak{Im}f_n(x) = 0` for all :math:`x\in (-1,0)`. Then :math:`\lim_{n\rightarrow \infty} \mathfrak{Im}f_n(1/2) = 0`. (b) Give a complete statement of the convergence theorem that you use in part (a). .. index:: residue theorem .. _2007Apr_3: .. proof:prob:: Use the residue theorem to evaluate :math:`\int_{-\infty}^{\infty}\frac{1}{1+x^4} dx`. .. _2007Apr_4: .. proof:prob:: Present a function :math:`f` that has all of the following properties: (i) :math:`f` is one-to-one and holomorphic on :math:`D`. (ii) :math:`\{f(z): z\in D\} = \{w\in \mathbb{C}: \mathfrak{Re}(w) > 0 \mbox{ and } \mathfrak{Im}(w) > 0\}`. (iii) :math:`f(0) = 1+i`. .. index:: maximum modulus theorem, Schwarz's lemma .. _2007Apr_5: .. proof:prob:: (a) Prove: If :math:`f: D\rightarrow D` is holomorphic and :math:`f(1/2) = 0`, then :math:`|f(0)| \leq 1/2`. (b) Give a complete statement of the maximum modulus theorem that you use in part (i). .. _2007Apr_6: .. proof:prob:: Prove: If :math:`G` is a connected open subset of :math:`\mathbb{C}`, any two points of :math:`G` can be connected by a parametric curve in :math:`G`. ----------------------------------- Solutions ---------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2007Apr_1>`. .. math:: f(z) = \frac{1}{z(z-1)} = \frac{1-z+z}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}. Let :math:`u=z+2`. Then :math:`z = u-2` and :math:`A = \{u\in \mathbb{C}: 2 < |u| < 3\}`. Therefore, .. math:: \frac{1}{z} = \frac{1}{u-2} = \frac{1}{u}\frac{1}{(1-2/u)} = \frac{1}{u}\sum_{n=0}^\infty \left(\frac{2}{u}\right)^n converges for :math:`|u|>2` and, substituting :math:`u=z+2` in the last expression, we have .. math:: \frac{1}{z} = \frac{1}{z+2}\sum_{n=0}^\infty \left(\frac{2}{z+2}\right)^n = \sum_{n=0}^\infty 2^n(z+2)^{-n-1} = \sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{n+1}(z+2)^n, converging for :math:`2<|z+2|`. Next, consider that .. math:: \frac{1}{z-1} = \frac{1}{u-3} = \frac{-1}{3(1-u/3)} = \frac{-1}{3}\sum_{n=0}^\infty \left(\frac{u}{3}\right)^n converges for :math:`|u|<3` and, substituting :math:`u=z+2` in the last expression, we have .. math:: \frac{1}{z-1} = -\sum_{n=0}^\infty \left(\frac{1}{3}\right)^{n+1}(z+2)^n, converging for :math:`|z+2|<3`. Therefore, .. math:: f(z) = \frac{1}{z-1} - \frac{1}{z} = -\sum_{n=0}^\infty \left(\frac{1}{3}\right)^{n+1}(z+2)^n-\sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{n+1}(z+2)^n, for :math:`z\in A`. .. ------------------------------------------------------------ .. .. container:: toggle .. .. container:: header Solution to :numref:`Problem {number} <2007Apr_2>`. (coming soon) .. ------------------------------------------------------------ .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2007Apr_3>`. Note that .. math:: f(z) = \frac{1}{1+x^4} =\frac{1}{(z^2 + i)(z^2 - i)}=\frac{1}{(z + e^{i\pi/4})(z - e^{i\pi/4})(z + e^{i3\pi/4})(z - e^{i3\pi/4})}, which reveals that the poles of :math:`f` in the upper half plane are at :math:`e^{i\pi/4}` and :math:`e^{i3\pi/4}`. Let :math:`\Gamma_R` be the contour shown in the figure below; i.e., :math:`\Gamma_R = g(R) \cup[-R,R]`, where :math:`R>1`. Then, by the residue theorem, .. math:: \int_{\Gamma_R} f(z) dz = 2\pi i \left[\mbox{Res}(f,e^{i\pi/4}) + \mbox{Res}(f,e^{i3\pi/4})\right]. :label: 3-1 .. todo:: add figure The other two poles of :math:`f` are in the lower half-plane, so both :math:`e^{i\pi/4}` and :math:`e^{i3\pi/4}` are simple poles. Therefore, .. math:: \mbox{Res}(f,e^{i\pi/4}) = \lim_{z\rightarrow e^{i\pi/4}} (z-e^{i\pi/4})f(z) = \frac{1}{2e^{i\pi/4}(e^{i\pi/4}-e^{i3\pi/4})(e^{i\pi/4}+e^{i3\pi/4})} = -\frac{1}{4} i e^{-i\pi/4}, .. math:: \mbox{Res}(f,e^{i3\pi/4}) = \lim_{z\rightarrow e^{i3\pi/4}} (z-e^{i3\pi/4})f(z) = \frac{1}{2e^{i3\pi/4}(e^{i3\pi/4}-e^{i\pi/4})(e^{i3\pi/4}+e^{i\pi/4})} = \frac{1}{4} i e^{-i3\pi/4}. Plugging these into :eq:`3-1` yields .. math:: \int_{\Gamma_R} f(z) dz = 2\pi i \left(\frac{1}{4} i e^{-i3\pi/4} -\frac{1}{4} i e^{-i\pi/4}\right) = \frac{\pi}{2}(e^{-i\pi/4} - e^{-i3\pi/4}) = \frac{\pi}{\sqrt{2}}. It remains to show .. math:: \lim_{R\rightarrow \infty} \left| \int_{g(R)} f(z) dz \right| = 0. Changing variables via :math:`z = Re^{i\theta} \; (0\leq \theta \leq \pi)`, .. math:: \left| \int_{g(R)} f(z) dz \right| = \left| \int_0^\pi \frac{i R e^{i\theta}}{1+(Re^{i\theta})^4}\right| \leq \frac{\pi R}{R^4 - 1} \rightarrow 0, \quad \text{as $R \rightarrow \infty$}. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2007Apr_4>`. First consider [1]_ :math:`\phi_1(z) = \frac{1-z}{1+z}`, which maps :math:`D` onto the right half-plane :math:`P^+= \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0\}`. Let :math:`\phi_2(z) = e^{i\pi/2}z = iz`, which maps :math:`P^+` onto the upper half-plane :math:`\Pi^+= \{z\in \mathbb{C}: \mathfrak{Im}(z) > 0\}`. Next, let :math:`\phi_3(z) = z^{1/2}` be a branch of the square root function on :math:`\Pi^+`. Then :math:`\phi_3` maps :math:`\Pi^+` onto the first quadrant :math:`Q_1 = \{z\in \mathbb{C}: 0 < \mbox{arg}(z) < \pi/2 \}`. The function :math:`\phi = \phi_3 \circ \phi_2 \circ \phi_1` satisfies the first two conditions, so we check whether it satisfies condition (iii): .. math:: \phi_1(0) = 1 \quad \Rightarrow \quad (\phi_2\circ \phi_1)(0) = \phi_2(1) = i \quad \Rightarrow \quad (\phi_3 \circ \phi_2 \circ \phi_1)(0) = \phi_3(i) = \frac{1+i}{\sqrt{2}} so apparently we're off by a factor of :math:`\sqrt{2}`. This is easy to fix: let :math:`\phi_4(z) = \sqrt{2} z`. Then the holomorphic function :math:`f \triangleq \phi_4 \circ \phi` maps :math:`D` bijectively onto :math:`Q_1` and :math:`f(0) = 1+i`, as desired. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2007Apr_5>`. (a) Define :math:`\phi(z) = \frac{1/2 - z}{1-z/2}`. This is a holomorphic bijection [2]_ of :math:`\bar{D}` onto :math:`\bar{D}`. Therefore, :math:`g = f \circ \phi \in H(D)`, :math:`|g(z)|\leq 1` for all :math:`z\in D`, and :math:`g(0) = f(\phi(0)) = f(1/2) = 0`. Thus :math:`g` satisfies the hypotheses of :ref:`Schwarz's lemma `, which allows us to conclude the following: (i) :math:`|g(z)| \leq |z|`, for all :math:`z\in D`, and (ii) :math:`|g'(0)| \leq 1`, with equality in (a) for some :math:`z\in D` or equality in (b) iff :math:`g(z) = e^{i\theta}z` for some constant :math:`\theta \in \mathbb R`. By condition (i), .. math:: 1/2 \geq |g(1/2)| = |f(\phi(1/2))| = |f(0)|. (b) In part (a) we used :ref:`Schwarz's lemma `. This is sometimes thought of as a version of the :ref:`maximum modulus principle `, since it is such an easy corollary of what is usually called the maximum modulus principle. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <2007Apr_6>`. First, recall that if :math:`A\subset G \subset \mathbb{C}`, then :math:`A` is said to be *open relative to* :math:`G`, or simply *open in* :math:`G`, if for any :math:`a\in A` there is a neighborhood :math:`B(a,\epsilon) = \{z\in \mathbb{C}: |z-a|<\epsilon\}` such that :math:`B(a,\epsilon)\cap G \subset A`. [3]_ Next, recall that a subset :math:`G\subset \mathbb{C}` is *connected* iff the only subsets of :math:`G` that are both open and closed relative to :math:`G` are the empty set and :math:`G` itself. Equivalently, if there exist non-empty disjoint subsets :math:`A, B \subset G` that are open in :math:`G` and have the property :math:`G = A\cup B`, then :math:`G` is not connected, or *disconnected*. [4]_ Now, suppose :math:`G` is a connected open subset of :math:`\mathbb{C}`. Fix :math:`z_0\in G` and let :math:`\Omega \subset G` be the subset of points that can be connected to :math:`z_0` by a parametric curve in :math:`G`. Since :math:`G` is open, :math:`\exists B(z_0,\epsilon) \subset G` for some :math:`\epsilon > 0`, and clearly :math:`B(z_0,\epsilon) \subset \Omega`. In particular, :math:`\Omega \neq \emptyset` . If we can show :math:`\Omega` is both open and closed in :math:`G`, then it will follow by connectedness that :math:`\Omega = G`, and the problem will be solved. (:math:`\Omega` is open) Let :math:`w\in \Omega` be connected to :math:`z_0` by a parametric curve :math:`\gamma \subset G`. Since :math:`G` is open, :math:`\exists \epsilon > 0` such that :math:`B(w,\epsilon)\subset G`. Clearly any :math:`w_1 \in B(w,\epsilon)` can be connected to :math:`z_0` by a parametric curve (from :math:`w_1` to :math:`w`, then from :math:`w` to :math:`z_0` via :math:`\gamma`) that remains in :math:`G`. This proves that :math:`B(w,\epsilon)\subset \Omega`, so :math:`\Omega` is open. (:math:`\Omega` is closed) We show :math:`G \setminus \Omega` is open (and thus, in fact, empty). If :math:`z \in G \setminus \Omega`, then, since :math:`G` is open, :math:`\exists \delta >0` such that :math:`B(z,\delta)\subset G`. We want :math:`B(z,\delta)\subset G \setminus \Omega`. This must be true since, otherwise, there would be a point :math:`z_1 \in B(z,\delta) \cap\Omega` which could be connected to both :math:`z` and :math:`z_0` by parametric curves in :math:`G`. But then a parametric curve in :math:`G` connecting :math:`z` to :math:`z_0` could be constructed, which would put :math:`z` in :math:`\Omega`---a contradiction. We have thus shown that :math:`\Omega` is both open and closed in :math:`G`, as well as non-empty. Since :math:`G` is connected, :math:`\Omega = G`. --------------------------------- .. rubric:: Footnotes .. [1] This is my favorite Möebius map. Not only does it map the unit disk onto the right half-plane, but also it maps the right half-plane onto the unit disk. This feature makes :math:`\phi_1` an extremely useful tool for conformal mapping problems, where you’re frequently required to map half-planes to the unit disk and vice-versa. Another nice feature of this map is that :math:`\phi_1^{-1} = \phi_1`. (Of course this must be the case if :math:`\phi_1` is to have the first feature.) Also note that, like all linear fractional transformations, :math:`\phi_1` is a holomorphic bijection of :math:`\mathbb{C}`. Therefore, if :math:`\phi_1` is to map the interior of the unit disk to the right half-plane, it must also map the exterior of the unit disk to the left half-plane. .. [2] See Rudin :cite:`Rudin:1987` Pages 254-5 (in particular, Theorem 12.4) for a nice discussion of functions of the form :math:`\phi_\alpha(z) = \frac{z-\alpha}{1-\bar{\alpha}z}`. In addition to 12.4, Sec. 12.5 and Theorem 12.6 are popular exam questions. .. [3] For example, the set :math:`A = [0,1]`, although closed in :math:`\mathbb{C}`, is open in :math:`G = [0,1]\cup \{2\}`. .. [4] To see the equivalence note that, in this case, :math:`A` is open in :math:`G`, as is :math:`A^c = G\setminus A = B`, so :math:`A` is both open and closed in :math:`G`. Also it is instructive to check, using either definition, that :math:`G = [0,1]\cup \{2\}` is disconnected. --------------------------------------- .. blank