.. 1989 April .. ========== Examination 1 ============== 1989 Apr -------- | **Instructions.** Do at least four problems. | **Time Limit.** 1.5 hours [1]_ .. index:: Liouville's theorem .. _1989Apr_1: .. proof:prob:: (a) Let :math:`U` be the unit disk in the complex plane :math:`\mathbb{C}`, :math:`U = \{z\in \mathbb{C}: |z| < 1 \}`, and let :math:`f` be an analytic function in a neighborhood of the closure of :math:`U`. Show that if :math:`f` is real on all the boundary of :math:`U`, then :math:`f` must be constant. (b) Let :math:`u` be a real harmonic function in all the complex plane :math:`\mathbb{C}`. Show that if :math:`u(z)\geq 0` for all :math:`z\in \mathbb{C}`,then :math:`u` must be constant. .. index:: Cauchy's formula .. ---------------- .. _1989Apr_2: .. proof:prob:: Let :math:`f` be an analytic function in the region :math:`\{z : |z|>1\}`, and suppose that .. math:: \lim_{z\rightarrow \infty} f(z) = 0. Show that if :math:`|z|>2`, then .. math:: \frac{1}{2\pi i} \int_{|\zeta| = 2} \frac{f(\zeta)}{\zeta - z} \, d\zeta = -f(z). :label: negf .. ---------------- .. index:: conformal mapping .. _1989Apr_3: .. proof:prob:: Let :math:`U` be the open unit disk in :math:`\mathbb{C}`, and let :math:`U^+` be the top half of this disk, .. math:: U^+ = \{z\in \mathbb{C}: \mathfrak{Im}(z) > 0, |z| < 1 \}. Exhibit a one-to-one conformal mapping from :math:`U^+` onto :math:`U`. .. ---------------- .. _1989Apr_4: .. proof:prob:: Let :math:`\{f_n\}` be a sequence of analytic functions in the unit disk :math:`U`, and suppose there exists a constant :math:`M` such that .. math:: \int_C |f_n(z)| \, |dz| \leq M for each :math:`f_n` and for all circles :math:`C` lying in :math:`U`. Prove that :math:`\{f_n\}` has a subsequence converging uniformly on compact subsets of :math:`U`. .. ---------------- .. _1989Apr_5: .. proof:prob:: Let :math:`Q` be a complex polynomial with distinct *simple* roots at the points :math:`a_1, a_2, \dots, a_n`, and let :math:`P` be a complex polynomial of degree less than that of :math:`Q`. Show that .. math:: \frac{P(z)}{Q(z)} = \sum_{k=1}^n \frac{P(a_k)}{Q'(a_k)(z-a_k)}. .. index:: residue theorem, contour integration .. ---------------- .. _1989Apr_6: .. proof:prob:: Use contour integration and the residue method to evaluate the integral .. math:: \int_0^\infty \frac{\cos x}{(1+x^2)^2} \, dx. -------------------------------------------- Solutions --------- .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <1989Apr_1>` (a) The hypotheses imply that :math:`\mathfrak{Im} f(e^{i\theta}) = 0` for all :math:`\theta\in \mathbb R`. Since :math:`f` is holomorphic in a neighborhood :math:`\Omega` of :math:`U`, the series .. math:: f(z) = \sum_{n=0}^\infty a_n z^n converges uniformly on any compact subset of :math:`\Omega`. The unit circle :math:`\Bbb{T}= \{z : |z|=1\} = \{e^{i\theta} : \theta \in \mathbb R\}` is one such compact subset, and here the series is .. math:: f(e^{i\theta}) = \sum_{n=0}^\infty a_n e^{i n\theta}. If we write the coefficients as :math:`a_n = c_n + ib_n`, where :math:`c_n, b_n \in \mathbb R`, then we have .. math:: 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 .. math:: \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 :math:`\theta \in [0, 2\pi]`. Therefore, with the possible exception of :math:`c_0`, we have :math:`c_n = b_n = 0`, for all :math:`n`, so :math:`f\equiv c_0`.   (b) Since :math:`\mathbb{C}` is simply connected, there is a real-valued harmonic conjugate :math:`v(z)` such that the function :math:`f(z) = u(z) + i v(z)` is entire. Now, since :math:`u(z)\geq 0`, :math:`f(z)` maps the complex plane into the right half-plane, :math:`\{\mathfrak{Re} f(z) \geq 0\}`. It follows immediately from Picard's theorem that :math:`f` must be constant. [2]_ However, an elementary argument using only :ref:`Liouville's theorem ` is probably preferable, so let :math:`f` be as above, and define :math:`g(z) = f(z) + 1`. Then :math:`g` is entire and maps :math:`\mathbb{C}` into :math:`\{w \in \mathbb{C}: \mathfrak{Re}(w) \geq 1\}`. In particular, :math:`g(z)` is bounded away from zero, so the function :math:`h(z) = 1/g(z)` is a bounded entire function. (In fact, :math:`|h(z)|\leq 1`.) Therefore, by :ref:`Liouville's theorem `, :math:`h` is constant, so :math:`f` is constant, so :math:`u = \mathfrak{Re} f` is constant. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <1989Apr_2>` By Cauchy's formula, if :math:`|z| |z| > 2 `. Fix :math:`\epsilon > 0`. Let :math:`R_\epsilon` be such that :math:`|f(\zeta)|< \epsilon/2` for all :math:`|\zeta| = R_\epsilon` and :math:`R_\epsilon > 2|z|`. Then :math:`|\zeta - z| > R_\epsilon/2` for all :math:`|\zeta| = R_\epsilon`, so .. math:: \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 :eq:`CF`, .. math:: \left|\frac{1}{2\pi i} \int_{|\zeta|=2} \frac{f(\zeta)}{\zeta - z} \, d\zeta + f(z)\right| < \epsilon. This holds for any :math:`\epsilon`, which proves :eq:`negf`. .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <1989Apr_3>` Consider :math:`\varphi_0(z) = \frac{1-z}{1+z}`, a linear fractional transformation which takes :math:`U` onto the right half-plane :math:`\Omega = \{z\in \mathbb{C}: \mathfrak{Re}(z) > 0\}`. (This property of :math:`\varphi_0` can be seen by considering :math:`\varphi_0(0) = 1` and :math:`\varphi_0(\infty) = -1` and arguing by symmetry.) Note that :math:`\varphi_0(1) = 0` and :math:`\varphi_0(x) \in \mathbb R`, for all :math:`x\in \mathbb R`. Also, :math:`\varphi_0` is conformal, so it preserves the right angle formed by the intersection of the circle and the real axis at the point :math:`z = 1`. Therefore, :math:`\varphi_0` takes :math:`U^+` onto either the first quadrant, :math:`\Omega^+ = \{z\in \Omega: \mathfrak{Im}(z) > 0\}`, or the fourth quadrant, :math:`\Omega^- = \{z\in \Omega: \mathfrak{Im}(z) < 0\}`. To see which, consider :math:`z = i/2`. .. math:: \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, :math:`\varphi_0: U^+ \rightarrow \Omega^-`. Let :math:`\varphi_1(z) = iz`, so that :math:`\varphi_1:\Omega^- \rightarrow \Omega^+`, let :math:`\varphi_2(z) = z^2`, so that :math:`\varphi_2:\Omega^+ \rightarrow \{\mathfrak{Im}(z) > 0\}`, and let :math:`\varphi_3(z) = -iz`, so that :math:`\varphi_3:\{\mathfrak{Im}(z) > 0\}\rightarrow \Omega = \{\mathfrak{Re}(z) > 0\}`. Finally, note that :math:`\varphi_0^{-1}(z) = \varphi_0(z)` maps :math:`\Omega` onto :math:`U`. Putting it all together, we see that a map satisfying the requirements is .. math:: \varphi(z) = (\varphi_0 \circ\varphi_3 \circ\varphi_2 \circ\varphi_1 \circ\varphi_0)(z). .. .. container:: toggle .. .. container:: header Solution to :numref:`Problem {number} <1989Apr_4>` (See the solution to :ref:`this problem <1991Nov_B3>` of the Nov 1991 exam.) .. container:: toggle .. container:: header Solution to :numref:`Problem {number} <1989Apr_5>` Denote the integral by :math:`I`. Since the integrand is even, .. math:: 2I = \int_{-\infty}^\infty \frac{\cos x}{(1+x^2)^2} \, dx. Consider the simple closed contour :math:`\Gamma_R = \gamma_R\cup [-R,R]`, where the trace of :math:`\gamma_R` is the set :math:`\{Re^{i\theta}: 1\leq \theta \leq \pi\}`, oriented counter-clockwise. Note that, if :math:`R>1`, then :math:`i` is inside the region bounded by :math:`\Gamma_R`. The function .. math:: f(z) = \frac{\cos z}{(1+z^2)^2} = \frac{\cos z}{(z+i)^2(z-i)^2} is holomorphic inside and on :math:`\Gamma_R`, except for a double pole at :math:`z = i`, where the residue is computed as follows: .. math:: \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} By the residue theorem, it follows that, for all :math:`R>1`, .. math:: \int_{\Gamma_R} f(z) \, dz = 2\pi i \, \mathrm{Res}(f,i) = \frac{\pi}{2e}. It remains to check that :math:`\int_{\gamma_R} f(z)\, dz \rightarrow 0`, as :math:`R\rightarrow \infty`, which will allow us to conclude that .. math:: 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}. :label: 3 Indeed, .. math:: \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 :math:`R>1`, so, letting :math:`R\rightarrow \infty`, we have :math:`\int_{\gamma_R} f(z)\, dz \rightarrow 0`. Therefore, by :eq:`3`, .. math:: I = \int_0^\infty \frac{\cos x}{(1+x^2)^2} \, dx = \frac{\pi}{4e}. .. .. container:: toggle .. .. container:: header Solution to :numref:`Problem {number} <1989Apr_6>` (coming soon) -------------------------------- .. rubric:: 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 :math:`\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. --------------------------------- .. insert space here