# Examination 1¶

## 1989 Apr¶

Instructions. Do at least four problems.
Time Limit. 1.5 hours 1
Problem 1
1. Let $$U$$ be the unit disk in the complex plane $$\mathbb{C}$$, $$U = \{z\in \mathbb{C}: |z| < 1 \}$$, and let $$f$$ be an analytic function in a neighborhood of the closure of $$U$$. Show that if $$f$$ is real on all the boundary of $$U$$, then $$f$$ must be constant.

2. Let $$u$$ be a real harmonic function in all the complex plane $$\mathbb{C}$$. Show that if $$u(z)\geq 0$$ for all $$z\in \mathbb{C}$$,then $$u$$ must be constant.

Problem 2

Let $$f$$ be an analytic function in the region $$\{z : |z|>1\}$$, and suppose that

$\lim_{z\rightarrow \infty} f(z) = 0.$

Show that if $$|z|>2$$, then

(1)$\frac{1}{2\pi i} \int_{|\zeta| = 2} \frac{f(\zeta)}{\zeta - z} \, d\zeta = -f(z).$
Problem 3

Let $$U$$ be the open unit disk in $$\mathbb{C}$$, and let $$U^+$$ be the top half of this disk,

$U^+ = \{z\in \mathbb{C}: \mathfrak{Im}(z) > 0, |z| < 1 \}.$

Exhibit a one-to-one conformal mapping from $$U^+$$ onto $$U$$.

Problem 4

Let $$\{f_n\}$$ be a sequence of analytic functions in the unit disk $$U$$, and suppose there exists a constant $$M$$ such that

$\int_C |f_n(z)| \, |dz| \leq M$

for each $$f_n$$ and for all circles $$C$$ lying in $$U$$. Prove that $$\{f_n\}$$ has a subsequence converging uniformly on compact subsets of $$U$$.

Problem 5

Let $$Q$$ be a complex polynomial with distinct simple roots at the points $$a_1, a_2, \dots, a_n$$, and let $$P$$ be a complex polynomial of degree less than that of $$Q$$. Show that

$\frac{P(z)}{Q(z)} = \sum_{k=1}^n \frac{P(a_k)}{Q'(a_k)(z-a_k)}.$
Problem 6

Use contour integration and the residue method to evaluate the integral

$\int_0^\infty \frac{\cos x}{(1+x^2)^2} \, dx.$

## Solutions¶

Solution to Problem 1

1. The hypotheses imply that $$\mathfrak{Im} f(e^{i\theta}) = 0$$ for all $$\theta\in \mathbb R$$. Since $$f$$ is holomorphic in a neighborhood $$\Omega$$ of $$U$$, the series

$f(z) = \sum_{n=0}^\infty a_n z^n$

converges uniformly on any compact subset of $$\Omega$$. The unit circle $$\Bbb{T}= \{z : |z|=1\} = \{e^{i\theta} : \theta \in \mathbb R\}$$ is one such compact subset, and here the series is

$f(e^{i\theta}) = \sum_{n=0}^\infty a_n e^{i n\theta}.$

If we write the coefficients as $$a_n = c_n + ib_n$$, where $$c_n, b_n \in \mathbb R$$, then we have

$a_n e^{i n \theta} = (c_n + ib_n)[\cos (n\theta) + i \sin (n\theta)] = [c_n \cos(n\theta)- b_n \sin(n\theta)] + i[c_n \sin(n\theta) + b_n \cos(n\theta)].$

Thus, by the hypothesis, the series

$\mathfrak{Im} f(e^{i\theta}) = \sum_{n=0}^\infty [c_n \sin(n\theta) + b_n \cos(n\theta)]$

converges uniformly to zero for all $$\theta \in [0, 2\pi]$$. Therefore, with the possible exception of $$c_0$$, we have $$c_n = b_n = 0$$, for all $$n$$, so $$f\equiv c_0$$.

2. Since $$\mathbb{C}$$ is simply connected, there is a real-valued harmonic conjugate $$v(z)$$ such that the function $$f(z) = u(z) + i v(z)$$ is entire.

Now, since $$u(z)\geq 0$$, $$f(z)$$ maps the complex plane into the right half-plane, $$\{\mathfrak{Re} f(z) \geq 0\}$$. It follows immediately from Picard’s theorem that $$f$$ must be constant. 2

However, an elementary argument using only Liouville’s theorem is probably preferable, so let $$f$$ be as above, and define $$g(z) = f(z) + 1$$.

Then $$g$$ is entire and maps $$\mathbb{C}$$ into $$\{w \in \mathbb{C}: \mathfrak{Re}(w) \geq 1\}$$. In particular, $$g(z)$$ is bounded away from zero, so the function $$h(z) = 1/g(z)$$ is a bounded entire function. (In fact, $$|h(z)|\leq 1$$.)

Therefore, by Liouville’s theorem, $$h$$ is constant, so $$f$$ is constant, so $$u = \mathfrak{Re} f$$ is constant.

Solution to Problem 2

By Cauchy’s formula, if $$|z| <R$$, then

(2)$f(z) = \frac{1}{2\pi i} \int_{|\zeta|=R} \frac{f(\zeta)}{\zeta - z} \, d\zeta - \frac{1}{2\pi i} \int_{|\zeta|=2} \frac{f(\zeta)}{\zeta - z} \, d\zeta.$

Note that this holds for all $$R > |z| > 2$$epsilon > 0`. Let $$R_\epsilon$$ be such that $$|f(\zeta)|< \epsilon/2$$ for all $$|\zeta| = R_\epsilon$$ and $$R_\epsilon > 2|z|$$. Then $$|\zeta - z| > R_\epsilon/2$$ for all $$|\zeta| = R_\epsilon$$, so

$\frac{1}{2\pi} \int_{|\zeta|=R_\epsilon} \frac{|f(\zeta)|}{|\zeta - z|} \, |d\zeta| < \frac{\epsilon/2}{2\pi} \, \frac{2\pi R_\epsilon}{R_\epsilon/2} = \epsilon.$

Therefore, by (2),

$\left|\frac{1}{2\pi i} \int_{|\zeta|=2} \frac{f(\zeta)}{\zeta - z} \, d\zeta + f(z)\right| < \epsilon.$

This holds for any $$\epsilon$$, which proves (1).

Solution to Problem 3

Consider $$\varphi_0(z) = \frac{1-z}{1+z}$$, a linear fractional transformation which takes $$U$$ onto the right half-plane $$\Omega = \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0\}$$. (This property of $$\varphi_0$$ can be seen by considering $$\varphi_0(0) = 1$$ and $$\varphi_0(\infty) = -1$$ and arguing by symmetry.)

Note that $$\varphi_0(1) = 0$$ and $$\varphi_0(x) \in \mathbb R$$, for all $$x\in \mathbb R$$. Also, $$\varphi_0$$ is conformal, so it preserves the right angle formed by the intersection of the circle and the real axis at the point $$z = 1$$. Therefore, $$\varphi_0$$ takes $$U^+$$ onto either the first quadrant, $$\Omega^+ = \{z\in \Omega: \mathfrak{Im}(z) > 0\}$$, or the fourth quadrant, $$\Omega^- = \{z\in \Omega: \mathfrak{Im}(z) < 0\}$$. To see which, consider $$z = i/2$$.

$\varphi_0(i/2) = \frac{1-i/2}{1+i/2}= \left(\frac{1-i/2}{1+i/2}\right)\left(\frac{1-i/2}{1-i/2}\right) = \frac{3-4i}{5} \in \Omega^-.$

Thus, $$\varphi_0: U^+ \rightarrow \Omega^-$$.

Let $$\varphi_1(z) = iz$$, so that $$\varphi_1:\Omega^- \rightarrow \Omega^+$$, let $$\varphi_2(z) = z^2$$, so that $$\varphi_2:\Omega^+ \rightarrow \{\mathfrak{Im}(z) > 0\}$$, and let $$\varphi_3(z) = -iz$$, so that $$\varphi_3:\{\mathfrak{Im}(z) > 0\}\rightarrow \Omega = \{\mathfrak{Re}(z) > 0\}$$. Finally, note that $$\varphi_0^{-1}(z) = \varphi_0(z)$$ maps $$\Omega$$ onto $$U$$. Putting it all together, we see that a map satisfying the requirements is

$\varphi(z) = (\varphi_0 \circ\varphi_3 \circ\varphi_2 \circ\varphi_1 \circ\varphi_0)(z).$

Solution to Problem 4

(See the solution to this problem of the Nov 1991 exam.)

Solution to Problem 5

Denote the integral by $$I$$. Since the integrand is even,

$2I = \int_{-\infty}^\infty \frac{\cos x}{(1+x^2)^2} \, dx.$

Consider the simple closed contour $$\Gamma_R = \gamma_R\cup [-R,R]$$, where the trace of $$\gamma_R$$ is the set $$\{Re^{i\theta}: 1\leq \theta \leq \pi\}$$, oriented counter-clockwise. Note that, if $$R>1$$, then $$i$$ is inside the region bounded by $$\Gamma_R$$.

The function

$f(z) = \frac{\cos z}{(1+z^2)^2} = \frac{\cos z}{(z+i)^2(z-i)^2}$

is holomorphic inside and on $$\Gamma_R$$, except for a double pole at $$z = i$$, where the residue is computed as follows:

\begin{split}\begin{aligned} \mathrm{Res}(f,i) &= \lim_{z\rightarrow i} \frac{d}{dz}[(z-i)^2 f(z)] = \lim_{z\rightarrow i} \frac{d}{dz}\frac{\cos z}{(z+i)^2}\\[4pt] &=\lim_{z\rightarrow i} \frac{-(z+i)^2\sin z - 2(z+i) \cos z }{(z+i)^4}\\[4pt] &=\frac{-(2i)^2\sin(i) - 4i \cos(i)}{(2i)^4}\\[4pt] &=\frac{\sin(i)- i \cos(i)}{4}=\frac{-ie^{i\cdot i}}{4} =\frac{-i}{4e}. \end{aligned}\end{split}

By the residue theorem, it follows that, for all $$R>1$$,

$\int_{\Gamma_R} f(z) \, dz = 2\pi i \, \mathrm{Res}(f,i) = \frac{\pi}{2e}.$

It remains to check that $$\int_{\gamma_R} f(z)\, dz \rightarrow 0$$, as $$R\rightarrow \infty$$, which will allow us to conclude that

(3)$2I = \int_{-\infty}^{\infty} f(x) \, dx = \lim_{R\rightarrow \infty}\left[\int_{\Gamma_R} f(z)\,dz-\int_{\gamma_R} f(z)\,dz\right] = \lim_{R\rightarrow \infty}\int_{\Gamma_R} f(z)\,dz = \frac{\pi}{2e}.$

Indeed,

$\left|\int_{\gamma_R} f(z)\, dz\right| = \left|\int_{\gamma_R} \frac{\cos z}{(1+z^2)^2} \, dz \right|\leq \frac{1}{(R^2-1)^2}\ell(\gamma_R) = \frac{\pi R}{(R^2-1)^2}.$

This inequality holds for all $$R>1$$, so, letting $$R\rightarrow \infty$$, we have $$\int_{\gamma_R} f(z)\, dz \rightarrow 0$$. Therefore, by (3),

$I = \int_0^\infty \frac{\cos x}{(1+x^2)^2} \, dx = \frac{\pi}{4e}.$

Solution to Problem 6 (coming soon)

Footnotes

1

In 1989 there was a single three hour test covering both real and complex analysis. Students were required to do nine problems, with at least four from each part.

2

Picard’s theorem states that a non-constant entire function can omit at most one value of $$\mathbb{C}$$ from its range. This is a very powerful theorem and using it for an easy problem like this one is like killing a fly with a sledge hammer.

Real Analysis Exams

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