Examination 8

2007 Apr

Notation. \(\mathbb{C}\) is the set of complex numbers, \(D= \{z\in \mathbb{C}: |z|<1\}\), and, for any open set \(G\subset \mathbb{C}\), \(H(G)\) is the set of holomorphic functions on \(G\).

Problem 47

Give the Laurent series expansion of \(\frac{1}{z(z-1)}\) in the region \(A \equiv \{z\in \mathbb{C}: 2< |z+2| < 3\}\).

Problem 48
  1. Prove: Suppose that for all \(z \in D\) and all \(n\in \mathbb{N}\) we have that \(f_n\) is holomorphic in \(D\) and \(|f_n(z)|<1\). Also suppose that \(\lim_{n\rightarrow \infty} \mathfrak{Im}f_n(x) = 0\) for all \(x\in (-1,0)\). Then \(\lim_{n\rightarrow \infty} \mathfrak{Im}f_n(1/2) = 0\).

  2. Give a complete statement of the convergence theorem that you use in part (a).

Problem 49

Use the residue theorem to evaluate \(\int_{-\infty}^{\infty}\frac{1}{1+x^4} dx\).

Problem 50

Present a function \(f\) that has all of the following properties:

  1. \(f\) is one-to-one and holomorphic on \(D\).

  2. \(\{f(z): z\in D\} = \{w\in \mathbb{C}: \mathfrak{Re}(w) > 0 \mbox{ and } \mathfrak{Im}(w) > 0\}\).

  3. \(f(0) = 1+i\).

Problem 51
  1. Prove: If \(f: D\rightarrow D\) is holomorphic and \(f(1/2) = 0\), then \(|f(0)| \leq 1/2\).

  2. Give a complete statement of the maximum modulus theorem that you use in part (i).

Problem 52

Prove: If \(G\) is a connected open subset of \(\mathbb{C}\), any two points of \(G\) can be connected by a parametric curve in \(G\).


Solution to Problem 47.

\[f(z) = \frac{1}{z(z-1)} = \frac{1-z+z}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}.\]

Let \(u=z+2\). Then \(z = u-2\) and \(A = \{u\in \mathbb{C}: 2 < |u| < 3\}\). Therefore,

\[\frac{1}{z} = \frac{1}{u-2} = \frac{1}{u}\frac{1}{(1-2/u)} = \frac{1}{u}\sum_{n=0}^\infty \left(\frac{2}{u}\right)^n\]

converges for \(|u|>2\) and, substituting \(u=z+2\) in the last expression, we have

\[\frac{1}{z} = \frac{1}{z+2}\sum_{n=0}^\infty \left(\frac{2}{z+2}\right)^n = \sum_{n=0}^\infty 2^n(z+2)^{-n-1} = \sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{n+1}(z+2)^n,\]

converging for \(2<|z+2|\). Next, consider that

\[\frac{1}{z-1} = \frac{1}{u-3} = \frac{-1}{3(1-u/3)} = \frac{-1}{3}\sum_{n=0}^\infty \left(\frac{u}{3}\right)^n\]

converges for \(|u|<3\) and, substituting \(u=z+2\) in the last expression, we have

\[\frac{1}{z-1} = -\sum_{n=0}^\infty \left(\frac{1}{3}\right)^{n+1}(z+2)^n,\]

converging for \(|z+2|<3\). Therefore,

\[f(z) = \frac{1}{z-1} - \frac{1}{z} = -\sum_{n=0}^\infty \left(\frac{1}{3}\right)^{n+1}(z+2)^n-\sum_{n=-\infty}^{-1}\left(\frac{1}{2}\right)^{n+1}(z+2)^n,\]

for \(z\in A\).

Solution to Problem 48. (coming soon)

Solution to Problem 49.

Note that

\[f(z) = \frac{1}{1+x^4} =\frac{1}{(z^2 + i)(z^2 - i)}=\frac{1}{(z + e^{i\pi/4})(z - e^{i\pi/4})(z + e^{i3\pi/4})(z - e^{i3\pi/4})},\]

which reveals that the poles of \(f\) in the upper half plane are at \(e^{i\pi/4}\) and \(e^{i3\pi/4}\). Let \(\Gamma_R\) be the contour shown in the figure below; i.e., \(\Gamma_R = g(R) \cup[-R,R]\), where \(R>1\). Then, by the residue theorem,

(21)\[\int_{\Gamma_R} f(z) dz = 2\pi i \left[\mbox{Res}(f,e^{i\pi/4}) + \mbox{Res}(f,e^{i3\pi/4})\right].\]


add figure

The other two poles of \(f\) are in the lower half-plane, so both \(e^{i\pi/4}\) and \(e^{i3\pi/4}\) are simple poles. Therefore,

\[\mbox{Res}(f,e^{i\pi/4}) = \lim_{z\rightarrow e^{i\pi/4}} (z-e^{i\pi/4})f(z) = \frac{1}{2e^{i\pi/4}(e^{i\pi/4}-e^{i3\pi/4})(e^{i\pi/4}+e^{i3\pi/4})} = -\frac{1}{4} i e^{-i\pi/4},\]
\[\mbox{Res}(f,e^{i3\pi/4}) = \lim_{z\rightarrow e^{i3\pi/4}} (z-e^{i3\pi/4})f(z) = \frac{1}{2e^{i3\pi/4}(e^{i3\pi/4}-e^{i\pi/4})(e^{i3\pi/4}+e^{i\pi/4})} = \frac{1}{4} i e^{-i3\pi/4}.\]

Plugging these into (21) yields

\[\int_{\Gamma_R} f(z) dz = 2\pi i \left(\frac{1}{4} i e^{-i3\pi/4} -\frac{1}{4} i e^{-i\pi/4}\right) = \frac{\pi}{2}(e^{-i\pi/4} - e^{-i3\pi/4}) = \frac{\pi}{\sqrt{2}}.\]

It remains to show

\[\lim_{R\rightarrow \infty} \left| \int_{g(R)} f(z) dz \right| = 0.\]

Changing variables via \(z = Re^{i\theta} \; (0\leq \theta \leq \pi)\),

\[\left| \int_{g(R)} f(z) dz \right| = \left| \int_0^\pi \frac{i R e^{i\theta}}{1+(Re^{i\theta})^4}\right| \leq \frac{\pi R}{R^4 - 1} \rightarrow 0, \quad \text{as $R \rightarrow \infty$}.\]

Solution to Problem 50.

First consider 1 \(\phi_1(z) = \frac{1-z}{1+z}\), which maps \(D\) onto the right half-plane \(P^+= \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0\}\).

Let \(\phi_2(z) = e^{i\pi/2}z = iz\), which maps \(P^+\) onto the upper half-plane \(\Pi^+= \{z\in \mathbb{C}: \mathfrak{Im}(z) > 0\}\).

Next, let \(\phi_3(z) = z^{1/2}\) be a branch of the square root function on \(\Pi^+\). Then \(\phi_3\) maps \(\Pi^+\) onto the first quadrant \(Q_1 = \{z\in \mathbb{C}: 0 < \mbox{arg}(z) < \pi/2 \}\).

The function \(\phi = \phi_3 \circ \phi_2 \circ \phi_1\) satisfies the first two conditions, so we check whether it satisfies condition (iii):

\[\phi_1(0) = 1 \quad \Rightarrow \quad (\phi_2\circ \phi_1)(0) = \phi_2(1) = i \quad \Rightarrow \quad (\phi_3 \circ \phi_2 \circ \phi_1)(0) = \phi_3(i) = \frac{1+i}{\sqrt{2}}\]

so apparently we’re off by a factor of \(\sqrt{2}\). This is easy to fix: let \(\phi_4(z) = \sqrt{2} z\). Then the holomorphic function \(f \triangleq \phi_4 \circ \phi\) maps \(D\) bijectively onto \(Q_1\) and \(f(0) = 1+i\), as desired.

Solution to Problem 51.

  1. Define \(\phi(z) = \frac{1/2 - z}{1-z/2}\). This is a holomorphic bijection 2 of \(\bar{D}\) onto \(\bar{D}\). Therefore, \(g = f \circ \phi \in H(D)\), \(|g(z)|\leq 1\) for all \(z\in D\), and \(g(0) = f(\phi(0)) = f(1/2) = 0\). Thus \(g\) satisfies the hypotheses of Schwarz’s lemma, which allows us to conclude the following:

    1. \(|g(z)| \leq |z|\), for all \(z\in D\), and

    2. \(|g'(0)| \leq 1\), with equality in (a) for some \(z\in D\) or equality in (b) iff \(g(z) = e^{i\theta}z\) for some constant \(\theta \in \mathbb R\).

    By condition (i),

    \[1/2 \geq |g(1/2)| = |f(\phi(1/2))| = |f(0)|.\]
  2. In part (a) we used Schwarz’s lemma. This is sometimes thought of as a version of the maximum modulus principle, since it is such an easy corollary of what is usually called the maximum modulus principle.

Solution to Problem 52.

First, recall that if \(A\subset G \subset \mathbb{C}\), then \(A\) is said to be open relative to \(G\), or simply open in \(G\), if for any \(a\in A\) there is a neighborhood \(B(a,\epsilon) = \{z\in \mathbb{C}: |z-a|<\epsilon\}\) such that \(B(a,\epsilon)\cap G \subset A\). 3

Next, recall that a subset \(G\subset \mathbb{C}\) is connected iff the only subsets of \(G\) that are both open and closed relative to \(G\) are the empty set and \(G\) itself. Equivalently, if there exist non-empty disjoint subsets \(A, B \subset G\) that are open in \(G\) and have the property \(G = A\cup B\), then \(G\) is not connected, or disconnected. 4

Now, suppose \(G\) is a connected open subset of \(\mathbb{C}\). Fix \(z_0\in G\) and let \(\Omega \subset G\) be the subset of points that can be connected to \(z_0\) by a parametric curve in \(G\). Since \(G\) is open, \(\exists B(z_0,\epsilon) \subset G\) for some \(\epsilon > 0\), and clearly \(B(z_0,\epsilon) \subset \Omega\). In particular, \(\Omega \neq \emptyset\) . If we can show \(\Omega\) is both open and closed in \(G\), then it will follow by connectedness that \(\Omega = G\), and the problem will be solved.

(\(\Omega\) is open) Let \(w\in \Omega\) be connected to \(z_0\) by a parametric curve \(\gamma \subset G\). Since \(G\) is open, \(\exists \epsilon > 0\) such that \(B(w,\epsilon)\subset G\). Clearly any \(w_1 \in B(w,\epsilon)\) can be connected to \(z_0\) by a parametric curve (from \(w_1\) to \(w\), then from \(w\) to \(z_0\) via \(\gamma\)) that remains in \(G\). This proves that \(B(w,\epsilon)\subset \Omega\), so \(\Omega\) is open.

(\(\Omega\) is closed) We show \(G \setminus \Omega\) is open (and thus, in fact, empty). If \(z \in G \setminus \Omega\), then, since \(G\) is open, \(\exists \delta >0\) such that \(B(z,\delta)\subset G\). We want \(B(z,\delta)\subset G \setminus \Omega\). This must be true since, otherwise, there would be a point \(z_1 \in B(z,\delta) \cap\Omega\) which could be connected to both \(z\) and \(z_0\) by parametric curves in \(G\). But then a parametric curve in \(G\) connecting \(z\) to \(z_0\) could be constructed, which would put \(z\) in \(\Omega\)—a contradiction.

We have thus shown that \(\Omega\) is both open and closed in \(G\), as well as non-empty. Since \(G\) is connected, \(\Omega = G\).



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 \(\phi_1\) 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 \(\phi_1^{-1} = \phi_1\). (Of course this must be the case if \(\phi_1\) is to have the first feature.) Also note that, like all linear fractional transformations, \(\phi_1\) is a holomorphic bijection of \(\mathbb{C}\). Therefore, if \(\phi_1\) 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.


See Rudin [Rud87] Pages 254-5 (in particular, Theorem 12.4) for a nice discussion of functions of the form \(\phi_\alpha(z) = \frac{z-\alpha}{1-\bar{\alpha}z}\). In addition to 12.4, Sec. 12.5 and Theorem 12.6 are popular exam questions.


For example, the set \(A = [0,1]\), although closed in \(\mathbb{C}\), is open in \(G = [0,1]\cup \{2\}\).


To see the equivalence note that, in this case, \(A\) is open in \(G\), as is \(A^c = G\setminus A = B\), so \(A\) is both open and closed in \(G\). Also it is instructive to check, using either definition, that \(G = [0,1]\cup \{2\}\) is disconnected.

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Real Analysis Exams

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