Difference between revisions of "MA598C 2014 mockquals soln2" - Rhea

Latest revision as of 04:43, 10 August 2014

Post solutions for mock qual #2 here.  Please indicate authorship!

Problem 1

Problem 2

Suppose $ u: \mathbb C \to \mathbb R $ is a non-constant harmonic function. Show that the zero set $ S = \{z \in \mathbb C | u(z) = 0\} $ is unbounded as a subset of $ \mathbb C $.

Clinton, 2014

Suppose for contradiction that $ S $ is bounded; that is, $ \exists R \forall z $ such that $ |z| \geq R \implies u(z) \neq 0 $. Let $ f = u + iv $ analytic, where $ v $ is the global analytic conjugate for $ u $. We will show that $ g = e^{f(z)} $ is constant, thus $ u $ is constant.

Let $ z_0 = R+0i $. Then as $ |z_0|=R \geq R $, $ u(z_0) \neq 0 $. Without loss of generality (as we could multiply $ u,f $ by $ -1 $) let $ u(z_0) < 0 $. Consider a point $ p $ with $ |p|\geq R $. Let $ \gamma_p $ be the path along $ C_{|p|} $ clockwise to the origin, followed by the path along the real axis from $ |p|+0i $ to $ |R| $. This path lies outside $ B_R(0) $, and thus $ u(z) \neq 0 $ on this path; so by the contrapositive to the intermediate value theorem, as $ u $ is harmonic and thus continuous, $ u(z) < 0 $ at $ p $. Thus $ u(z) < 0 \forall z \in \mathbb C - D_R(0) $.

Consider now the analytic function $ g = e^{u+iv} = e^ue^{iv} $. For $ z \in \mathbb C -D_R(0) $, $ |g(z)| = |e^u||e^{iv}| = |e^u| < 1 $ as $ u(z)<0 $. On $ \overline{D_R(0)} $, as $ g $ is continuous on a compact set, it achieves a maximum $ M $. Thus $ g $ is a bounded entire function, so by Liouville, $ g $ is constant.

$ 0 \equiv g' \equiv f' e^f $. As $ e^f $ is nonzero, $ f' \equiv 0 $ and thus $ u' \equiv 0 $; so $ u $ is constant, a contradiction.

Thus no nonconstant $ u $ with bounded zero set exists.

Problem 3

Construct a LFT between $ D_1(0) \cap \left\{x + iy \in \mathbb C : y > \dfrac{\sqrt 2}2\right\} $ onto $ D_1(0) $.

Clinton, 2014

We will construct it in three stages. Consider the LFT taking $ -\frac{\sqrt{2}}2+\frac{\sqrt{2}}2i \to 0 $, $ \frac{\sqrt{2}}2+\frac{\sqrt{2}}2i \to \infty $, $ \frac{\sqrt{2}}2i \to 1 $, given by

$ f_1(z) = -\dfrac{z+\frac{\sqrt{2}}2-\dfrac{\sqrt{2}}2i}{z-\frac{\sqrt{2}}2-\frac{\sqrt{2}}2i} $

Then $ f(i) = \frac{\sqrt{2}}2+\frac{\sqrt{2}}2i $, with the interior mapping to the first octant.

Next expand to the upper half plane via $ f_2(z) = z^4 $.

Finally, take the upper half plane to the unit disc via $ f_3(z) = \dfrac{z-i}{z+i} $.

Thus $ f = f_3 \circ f_2 \circ f_1 $ takes the given region to the unit disk.

Problem 4

Let $ \displaystyle f(z) = \sum_{n=0}^\infty x^{n!} $. Show the radius of convergence of this power series is $ 1 $. Let $ u $ be a root of unity. Show that $ \lim_{r \to 1^-} f(ru) = \infty $.

Clinton, 2014

Let $ \Omega_\epsilon = D_1(0) \cup D_\epsilon(1) $. Is there $ \epsilon > 0 $ and a meromorphic function $ F $ with $ F \equiv f $ on $ D_1(0) $? Explain.

Note that, for $ x \in D_1(0) $, $ |\sum_{n=0}^\infty x^{n!}| \leq \sum_{n=0}^\infty |x|^{n!} \leq \sum_{n=0}^\infty |x|^n = \frac{1}{1-|x|} $, so the series is absolutely convergent and thus convergent at each point. Thus $ R \geq 1 $. We will show, in the next portion, it diverges for at least one (countably infinitely many) values with $ |x|=1 $, showing $ R = 1 $.

Let $ u $ be a $ j $th root of unity. We intend to show that $ \displaystyle\lim_{r \to 1-}f(ru)=\infty $; that is, $ \forall M \exists \epsilon $ such that $ 1-\epsilon < x < 1 \implies f(xu) > M $. Fix such an $ M $.

Let $ g_k(x) = \sum_{n=0}^k (ux)^{n!} $. Notice that, for $ k>j $,

$ \displaystyle g_k(1) = \sum_{n=0}^{j-1} u^{n!}+ \sum_{j}^k (u^j)^{\frac {n!}{j}} = \sum_{n=0}^{j-1} u^{n!} + (k-j) \to_{k \to \infty} \infty $

And that similarly, for positive $ x $, for $ k>j $ the summands are positive real.

So there exists some $ k $ for which $ g_k(1) > 2M $. By continuity of polynomials, $ \exists \epsilon $ for which $ |r-1|< \epsilon \implies |g_k(1)-g_k(r)| < M $; that is, $ g_k(r) > 2M-M = M $.

As the summands are positive for $ n>j $, $ f(ur) > g_k(r)=M $ on this neighborhood.

Thus $ f(ur) \to \infty $ as desired.

There is no such merromorphic function; a merromorphic function must be continuous except on a discrete set, but the roots of unity are dense on the infinite subset $ C_1(0) \cap d_\epsilon(1) $ for all $ \epsilon $, and the function diverges at every root of unity.

Problem 5

Evaluate the integral $ \displaystyle\int_0^\infty\frac {\sqrt x}{x^2+1}dx $ using techniques from complex variables.

Clinton, 2014

Consider the contour $ \gamma_R $ consisting of the subset of the real line $ [-R,R] $ followed by the upper half circle $ C_R^+ $ of $ C_R(0) $, traversed counterclockwise.

$ \displaystyle \int_{\gamma_R} \dfrac{z^{1/2}}{z^2+1} = \int_0^R \frac{\sqrt{x}}{x^2+1} + \int_{C_R^+} \frac{z^{1/2}}{z^2+1} + \int_{0}^R \frac{i \sqrt x}{x^2+1} = (1+i)\int_0^R \frac{\sqrt{x}}{x^2+1} + \int_{C_R^+} \frac{z^{1/2}}{z^2+1} $

Note that $ \displaystyle \left|\int_{C_R^+} \frac{z^{1/2}}{z^2+1}\right| \leq \pi R \max_{z \in C_R^+}\left| \frac{z^{1/2}}{z^2+1} \right| $

Which, by the basic polynomial estimate, $ \leq \pi R \frac{R^{1/2}}{AR^2}= \frac{\pi}{A} R^{-\frac 12} $ for some $ A $ and sufficiently large $ R $. As the power of $ R $ is negative, the limit as $ R \to \infty $ is zero.

Therefore $ \displaystyle\int_0^\infty \frac{\sqrt x}{x^2+1}= \frac{1}{1+i} \lim_{R \to \infty} \int_{\gamma_R} \frac{z^{1/2}}{z^2+1} $

By the residue theorem, for $ R > 1 $, the integral is also equal to the residue at $ i $ (there are two simple poles at $ i,-i $, the single zeroes of the denominator). By a theorem in class, the residue at a simple pole is given by $ \operatorname{res}_a \frac {f(z)}{g(z)} = 2 \pi i\frac{f(a)}{g'(a)} $ , so

$ \displaystyle\operatorname{res}_{i} \frac{z^{1/2}}{z^2+1} = 2 \pi i \frac{\frac{1+i}{\sqrt{2}}}{2i} = (1+i)\frac{\pi}{\sqrt 2} $

So $ \displaystyle\int_0^\infty \frac{\sqrt x}{x^2+1}= \frac \pi {\sqrt 2} $

Problem 6

Suppose $ f $ is analytic on $ D_1(0) $ and $ |f(z)| < 1 $ for all $ z $ in the unit disk. Prove that if $ f(0) = a \neq 0 $, then $ f $ has no zeroes on $ D_{|a|}(0) $.

Clinton, 2014

Let $ \phi_a := \dfrac{z-a}{1-\bar a z} $, the automorphism of the unit disk $ D = D_1(0) $ taking $ a \to 0 $ and $ 0 \to -a $. Let $ g \equiv \phi_a \circ f $; then as $ f: D \to D $, $ \phi_a: D \to D $, we have $ g: D \to D $ with $ g(0) = \phi_a(a) = 0 $. So by the Schwarz lemma, $ g(z) \leq z $.

For contradiction suppose there exists $ b \in D_{|a|}(0) $ such that $ f(b) = 0 $. Then

$ |a| = |-a| = |\phi_a(0)| = |g(b)| \leq |b| < |a| $, a contradiction. Therefore no such zeroes exist.

Problem 7

Suppose $ f $ is a one-to-one entire function. Show that there exists $ a \in \mathbb C-\{0\}, b \in \mathbb C $ such that $ f(z) = az + b $.

Clinton, 2014

Consider $ g(z) \equiv f\left( \dfrac 1z \right) $. This function has a sinularity at $ 0 $; this singularity is either essential, removable, or a pole. We will next show it must be a pole.

Suppose for contradiction that it is essential. Then by Picard's theorem, it attains $ g(z)=1 $ at infinitely many points in $ \mathbb C $; however, $ \frac 1z: \mathbb C - \{0\} \to \mathbb C $ is one to one, and $ f $ is one to one, so their composition is one to one; this is a contradiction.

Suppose for contradiction that it is removable. Then $ \lim_{z \to 0}g(z) = L $ for some $ L $; so $ \forall \epsilon \exists \delta $ such that $ \left| \dfrac 1z \right| < \epsilon \implies |f(z)-L| < \epsilon $; e.g. $ \exists M = \dfrac 1\epsilon $ such that $ \forall |z|>M $, $ |f(z)|<|L|+|\epsilon| $.

Thus $ f $ is bounded and entire, so by Liouville's, it is constant. This again contradicts one-to-one.

Therefore the singularity is a pole, of some order $ n \in \mathbb N $. As discussed in class, if $ f(z) = \sum_0^\infty a_i z^i $, then $ f\left(\dfrac 1z\right) = \sum_{-\infty}^0 a_{-i} z^i $. As the singularity is a pole, only finitely many (the first $ n $) of the $ a_i $ can be nonzero. Thus, $ f = \sum_0^n a_i z^i $, a polynomial.

Next we need show $ n=1 $.

By the fundamental theorem of calculus, $ f $ has $ n $ zeroes counting multiplicity. If two are distinct, $ f $ is not one-to-one; so $ f(z) \equiv k(z-a)^n $. Suppose for contradiction $ n>1 $, and let $ \psi $ be an $ n $th root of unity. Then $ f(a+\psi)=k\psi^n=k=k(\psi^2)^n = f(a+\psi^2) $, a contradiction.

Thus $ n \leq 1 $. as $ n $ is the order of a pole, $ n \geq 1 $; so $ n=1 $, and $ f= ax+b $ for some $ a \neq 0,b $ in $ \mathbb C $.