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 $$G$$.

Solutions¶

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

Todo

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, $$\Omega = G$$.

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 $$\phi_1$$ 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.

2

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.

3

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

4

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.

Real Analysis Exams

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