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

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


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


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

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)



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.


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.

https://ws-na.amazon-adsystem.com/widgets/q?_encoding=UTF8&MarketPlace=US&ASIN=0070006571&ServiceVersion=20070822&ID=AsinImage&WS=1&Format=_SL140_&tag=typefunc-20 https://ws-na.amazon-adsystem.com/widgets/q?_encoding=UTF8&MarketPlace=US&ASIN=0387903283&ServiceVersion=20070822&ID=AsinImage&WS=1&Format=_SL140_&tag=typefunc-20 https://ws-na.amazon-adsystem.com/widgets/q?_encoding=UTF8&MarketPlace=US&ASIN=0070542341&ServiceVersion=20070822&ID=AsinImage&WS=1&Format=_SL140_&tag=typefunc-20 https://ws-na.amazon-adsystem.com/widgets/q?_encoding=UTF8&MarketPlace=US&ASIN=0821844288&ServiceVersion=20070822&ID=AsinImage&WS=1&Format=_SL140_&tag=typefunc-20 https://ws-na.amazon-adsystem.com/widgets/q?_encoding=UTF8&MarketPlace=US&ASIN=0691113858&ServiceVersion=20070822&ID=AsinImage&WS=1&Format=_SL140_&tag=typefunc-20

Real Analysis Exams

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