Examination 3¶

1991 Nov¶

Instructions. In each of sections A, B, and C, do all but one problem.

Time Limit. 2 hours

Section A¶

(Do 3 of the 4 problems.)

Problem 12

Where does the function

\[f(z) = z \mathfrak{Re}(z) + \bar{z} \mathfrak{Im}(z) + \bar{z}\]

have a complex derivative? Compute the derivative wherever it exists.

Problem 13

Prove that any nonconstant polynomial with complex coefficients has at least one root.
From (a) it follows that every nonconstant polynomial \(P\) has the factorization

\[P(z) = a \prod_{n=1}^N (z - \lambda_n),\]

where \(a\) and each root \(\lambda_n\) are complex constants. Prove that if \(P\) has only real coefficients, then \(P\) has a factorization

(4)¶\[P(z) = a \prod_{k=1}^K (z - r_k) \prod_{m=1}^M (z^2 - b_m z + c_m),\]

where a and each \(r_k, b_m, c_m\) are real constants.

Problem 14

Use complex residue methods to compute the integral

\[\int_0^\pi \frac{1}{5 + 3 \cos \theta} \, d\theta.\]

Problem 15

Explain how to map an infinite strip (i.e.,the region strictly between two parallel lines) onto the unit disk by a one-to-one conformal mapping.
Two circles lie outside one another except for common point of tangency. Explain how to map the region exterior to both circles (including the point at infinity) onto an infinite strip by a one-to-one conformal mapping.

Section B¶

(Do 3 of the 4 problems.)

Problem 16

Suppose that \(f\) is analytic in the annulus \(1< |z|<2\), and that there exists a sequence of polynomials converging to \(f\) uniformly on every compact subset of this annulus. Show that \(f\) has an analytic extension to all of the disk \(|z|<2\).

Problem 17

Let \(f\) be analytic in \(|z|<2\), with the only zeros of \(f\) being the distinct points \(a_1, a_2, \ldots, a_n\), of multiplicities \(m_1, m_2, \dots, m_n\), respectively, and with each \(a_j\) lying in the disk \(|z|<1\). Given that \(g\) is analytic in \(|z|<2\), what is

\[\int_{|z|=1} \frac{f'(z) g(z)}{f(z)} \, dz \; \text{ ?}\]

(Verify your answer.)

Problem 18

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

\[\int_C |f_n(z)|\,|dz| \leq M\]

for each \(f_n\) and for every circle \(C\) lying in \(D\). Prove that \(\{f_n\}\) has a subsequence converging uniformly on compact subsets of \(D\).

Problem 19

State and prove:

the mean value property for analytic functions
the maximum principle for analytic functions.

Section C¶

(Do 2 of the 3 problems.)

Problem 20

Let \(X\) be a Hausdorff topological space, let \(K\) be a compact subset of \(X\), and let \(x\) be a point of \(X\) not in \(K\). Show that there exist open sets \(U\) and \(V\) such that

\[K \subset U, \; x\in V, \; U\cap V = \emptyset.\]

Problem 21

A topological space \(X\) satisfies the second axiom of countability. Prove that every open cover of \(X\) has a countable subcover.

Problem 22

Let \(X\) be a topological space, and let \(U\) be a subset of \(X\).

Show that if an open set intersects the closure of \(Y\) then it intersects \(Y\).
Show that if \(Y\) is connected and if \(Y\subset Z \subset \bar{Y}\), then \(Z\) is connected.

Solutions¶

Writing \(f\) in terms of the real and imaginary parts of \(z = x+iy\), we have

\[\begin{split}f(x+iy) &= (x+iy)x + (x-iy) y + x-iy\\ &= x^2+ xy + x +i (xy -y^2 -y)\\ &= u(x,y) + i v(x,y),\end{split}\]

where \(u(x,y)=x^2+ xy + x\) and \(v(x,y) = xy -y^2 -y\) are the real and imaginary parts of \(f\). Therefore,

(5)¶\[\begin{split}u_x=2x+ y + 1 \quad v_y = x - 2y - 1\\ u_y =x \quad v_x = y.\end{split}\]

If \(f\) is holomorphic in some region, the Cauchy-Riemann equations (\(u_x = v_y\), \(u_y = -v_x\)) must hold there. By (5), this requires \(2x+ y + 1 = x - 2y - 1\) and \(x = -y\). Substituting the second equation into the first yields \(-y+1 = -3y -1\), or \(y=-1\). Then, since \(x = -y\), we must have \(x = 1\). Therefore, \(f\) has a complex derivative at \((x,y) = (1, -1)\), or \(z = 1-i\).

For any region \(\Omega \subseteq \mathbb{C}\), we define the linear functional \(\partial : H(\Omega) \rightarrow \mathbb{C}\) by \(\partial = \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}\right)\), and recall that, if \(f\in H(\Omega)\), then the derivative of \(f\) is given by \(f'(z) = (\partial f)(z)\), \(z\in \Omega\). In the present case,

\[\frac{\partial f}{\partial x} = 2x + y + 1 + iy, \quad \frac{\partial f}{\partial y} = x + i(x -2y - 1).\]

Therefore, \(\partial f(x+iy) = \frac{1}{2} [(2x + y + 1 + iy) -i(x + i(x -2y - 1))]= \frac{1}{2}[(3x-y) + i(y-x)]\), and finally,

\[f'(1-i) = \frac{1}{2}(4-2i) = 2-i.\]

Let \(p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n\), where \(\{a_i : 0\leq i \leq n\}\subseteq \mathbb C\).

Suppose \(p(z) \neq 0\) for all \(z\in \mathbb C\). Then consider the entire function \(f(z) = \frac{1}{p(z)}\).

Since \(\lim_{|z|\to \infty} |p(z)| = \infty\), it follows that, for every \(M>0\), there exists \(R>0\) such that

\[|z| \geq R \; \Rightarrow \; |p(z)|>M\]

On the other hand, for \(|z| \leq R\), we have, by the maximum modulus principle, that

\[|f(z)| \leq \max_{0\leq \theta \leq 2\pi} |f(Re^{i\theta})| < \frac{1}{M}.\]

Thus, \(f(z) = 1/p(z)\) is an entire bounded function, so Liouville’s Theorem implies \(f\) is constant, so \(p\) is constant, which is a contradiction.
Let \(P(z)\) be a polynomial of degree \(N\) with coefficients in \(\mathbb R\). Let \(\lambda_n\) be a complex root of \(P(z)\). Then \(0 = P(\lambda_n) = a_0 + a_1\lambda_n + \cdots + a_N \lambda_n^N\) implies \(P(\bar{\lambda_n}) = \bar{a_0} + \bar{a_0}\bar{\lambda_n} + \cdots + \bar{a_N}\bar{\lambda_n}^N = a_0 + a_0\bar{\lambda_n} + \cdots + a_N\bar{\lambda_n}^N\).

Therefore, \(\bar{\lambda_n}\) is also a root of \(P(z)\). It follows that \(P(z)\) has the factorization

(6)¶\[P(z) = a \prod_{k=1}^K (z - r_k) \prod_{m=1}^M (z- \lambda_m) \prod_{m = 1}^M (z - \bar{\lambda_n}),\]

where \(\{r_k\}\subseteq \mathbb R\) and \(\{\lambda_m\}\subseteq \mathbb C\).

Finally, from the calculation

\[(z-\lambda_m)(z - \bar{\lambda_m}) = z^2 - (\lambda_m + \hat{\lambda_m})z + |\lambda_n|^2 = z^2 - 2(\mathfrak{Re}\lambda_m)z + |\lambda_m|^2,\]

it is clear that (6) is of the same form as (4).

Let \(I = \int_0^\pi \frac{1}{5 + 3 \cos \theta} \, d\theta\). Note that \(\cos \theta\) is an even function (i.e.,\(\cos(-\theta) = \cos \theta\)), so

\[2I = \int_{-\pi}^\pi \frac{1}{5 + 3 \cos \theta} \, d\theta.\]

For \(z = e^{i\theta}\),

\[\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{1}{2}(z + \frac{1}{z}),\]

and \(dz = i e^{i\theta} d\theta\), from which it follows that

\[\begin{split}2I &= \int_{|z| = 1} \frac{1}{5 + \frac{3}{2}(z + \frac{1}{z})}\,\frac{dz}{iz}\\[6pt] &= \frac{1}{i}\int_{|z| = 1} \frac{dz}{5z + \frac{3}{2}(z^2 + 1)}\\[6pt] &= \frac{2}{3i}\int_{|z| = 1} \frac{dz}{z^2 + \frac{10}{3}z + 1}.\end{split}\]

Let \(p(z) = z^2 + \frac{10}{3}z + 1\). Then the roots of \(p(z)\) are \(z_1 = -1/3\) and \(z_2=-3\). Only \(z_1 = -1/3\) is inside the circle \(|z| = 1\), so the residue theorem implies

\[2I = \frac{2}{3i} \cdot 2\pi i \cdot \mathrm{Res}\left(\frac{1}{p(z)}, z_1\right).\]

Now,

\[\frac{1}{p(z)} = \frac{1}{(z-z_1)(z-z_2)},\]

which implies

\[\mathrm{Res}\left(\frac{1}{p(z)}, z_1\right) = \lim_{z\rightarrow z_1} \frac{1}{z-z_2} = \frac{1}{-\frac{1}{3} - (-3)} = \frac{3}{8}.\]

Therefore,

\[2I = \frac{2}{3i} \cdot 2\pi i \cdot \frac{3}{8} = \frac{\pi}{2},\]

so \(I = \frac{\pi}{4}\).

Solution to Problem 15 (coming soon)

Note that the function \(f\), being holomorphic in the annulus \(1<|z|<2\), has Laurent series representation

\[f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n,\]

converging locally uniformly for \(1< |z|<2\), where \(z_0\) is any point in the disk \(|z|<2\).

I claim that \(a_n = 0\) for all negative integers \(n\). To see this, first recall the formula for the coefficients in the Laurent series,

\[a_n = \frac{1}{2\pi i} \int_{|z|=R} \frac{f(z)}{(z-z_0)^{n+1}} \, dz, \quad (n\in \mathbb{Z};\, 1<R<2).\]

Let \(\{p_m\}\) be the sequence of polynomials mentioned in the problem statement.

Of course, \(p_m \in H(\mathbb{C})\), so Cauchy’s theorem implies \(\int_{|z|=R} p_m(z)\, dz = 0\), and, more generally,

\[\int_{|z|=R} p_m(z)(z-z_0)^{-n-1}\, dz = 0, \quad (n = -1, -2, \dots).\]

Therefore,

(7)¶\[\begin{split}|a_n| &= \frac{1}{2\pi} \left|\int_{|z|=R} \frac{f(z)}{(z-z_0)^{n+1}} \, dz- \int_{|z|=R} \frac{p_m(z)}{(z-z_0)^{n+1}}\, dz\right|\\[4pt] &\leq \frac{1}{2\pi} \int_{|z|=R} \frac{|f(z)-p_m(z)|}{|z-z_0|^{n+1}} \, |dz|.\end{split}\]

Finally, \(p_m \rightarrow f\) uniformly on \(|z| = R\), so (7) implies \(|a_n|=0\) for \(n = -1, -2, \dots\).

This proves that \(f(z) = \sum_{n=0}^\infty a_n (z-z_0)^n\), converging locally uniformly in \(|z|<2\). Whence \(f\in H(|z|<2)\).

Solution to Problem 17. (coming soon)

We must show that \(\mathcal{F}= \{f_n\}\) is a normal family.

If we can prove that \(\mathcal{F}\) is a locally bounded family of holomorphic functions—that is, \(\mathcal{F}\subset H(D)\) and, for any compact set \(K\subset D\), there is an \(M_K >0\) such that \(|f_n(z)| \leq M_K\) for all \(z\in K\) and all \(n=1,2,\dots\)—then the Montel theorem will give the desired result.

To show \(\mathcal{F}\) is locally bounded, it is equivalent to show that, for each point \(z_\alpha \in D\), there is a number \(M_\alpha\) and a neighborhood \(B(z_\alpha, r_\alpha)\subset D\) such that \(|f_n(z)|\leq M_\alpha\) for all \(z\in B(z_\alpha, r_\alpha)\) and all \(n=1,2, \ldots\). (Why is this equivalent?) 1

Fix \(z_\alpha \in D\). Let \(R_\alpha >0\) be such that \(\bar{B}(z_\alpha, R_\alpha) = \{z \in \mathbb{C}: |z-z_\alpha|\leq R_\alpha\}\subset D\). Then, for any \(z\in B(z_\alpha, R_\alpha/2)\), Cauchy’s formula gives

\[\begin{split}|f_n(z)| &\leq \frac{1}{2\pi} \int_{|\zeta - z_\alpha|=R_\alpha} \frac{|f_n(\zeta)|}{|\zeta - z|} \, |d\zeta|\\[4pt] &\leq \frac{1}{2\pi} \frac{1}{R_\alpha/2} \int_{|\zeta - z_\alpha|=R_\alpha} |f_n(\zeta)|\, |d\zeta|\\[4pt] &\leq \frac{M}{\pi R_\alpha}.\end{split}\]

The second inequality holds since \(|\zeta - z_\alpha|=R_\alpha\) and \(|z-z_\alpha|< R_\alpha/2\) imply \(|\zeta - z| > R_\alpha/2\). The last inequality follows from the hypothesis \(\int_C |f_n(z)|\,|dz| \leq M\) for any circle \(C\) in \(D\). Letting \(M_\alpha = \frac{M}{\pi R_\alpha}\), and \(r_\alpha = R_\alpha/2\), we have \(|f_n(z)| \leq M_\alpha\) for all \(z\in B(z_\alpha, r_\alpha)\) and all \(n=1,2,\dots\), as desired.

Theorem. (Mean-value property) Let \(G\subseteq \mathbb C\) be an open set containing a closed disk \(\bar{D}(a,r) = \{z : |z - a| \leq r\} \subseteq G\). Assume \(f\in H(G)\). Then

\[f(a) = \frac{1}{2\pi}\int_0^{2\pi} f(a+re^{i\theta})\, d\theta\]

Proof. By the Cauchy formula,

(8)¶\[f(a) = \frac{1}{2\pi i}\int_{|z-a|=r}\frac{f(z)}{z-a}\, dz\]

Let \(z = a + re^{i\theta}\). Then \(dz = ire^{i\theta}d\theta\) and (8) becomes

\[\begin{split}f(a) &= \frac{1}{2\pi i}\int_0^{2\pi}\frac{f(a+re^{i\theta})ire^{i\theta}}{a + re^{i\theta}-a}\, d\theta\\ &= \frac{1}{2\pi}\int_0^{2\pi}f(a+re^{i\theta})\, d\theta.\end{split}\]
Theorem. (Maximum modulus principle) Suppose \(G\subset \mathbb{C}\) is open and \(f\in H(G)\) attains its maximum modulus at some point \(a\in G\). Then \(f\) is constant.

In other words, if there exists \(a\in G\) such that \(|f(z)| \leq |f(a)|\) for all \(z\in G\), then \(f\) is constant.

Proof. Let \(\Omega = f(G)\). If \(f\) is not constant, then the open mapping theorem asserts that \(\Omega\) is open. In that case, we see that every \(f(a) = \omega \in \Omega\) is contained in a neighborhood \(B(\omega, \epsilon)\subseteq \Omega\), which implies that there are points \(z\in G\) such that \(|f(z)| > |\omega| = |f(a)|\). This completes the proof.

Solution to Problem 20 (coming soon)

Solution to Problem 21 (coming soon)

Solution to Problem 22 (coming soon)

Footnotes

1: Answer: If \(K\subset D\) is compact, we could select a finite covering of \(K\) by such neighborhoods \(B(z_{\alpha_j},r_{\alpha_j})\; (j=1, \dots, J)\) and then \(|f_n(z)|\leq \max_j M_{\alpha_j} \triangleq M_K\), for all \(z\in K\) and \(n=1,2,\dots\).