# Examination 1¶

## 1989 Apr¶

**Instructions.**Do at least four problems.

**Time Limit.**1.5 hours 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.

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.

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

Show that if \(|z|>2\), then

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

Exhibit a one-to-one conformal mapping from \(U^+\) onto \(U\).

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

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 \(U\).

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

Use contour integration and the residue method to evaluate the integral

## Solutions¶

Solution to Problem 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 \(f\equiv c_0\).

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

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

Therefore, by (2),

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 \(z = i/2\).

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

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,

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

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

By the residue theorem, it follows that, for all \(R>1\),

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

Indeed,

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),

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.

Please email comments, suggestions, and corrections to williamdemeo@gmail.com