Post solutions for mock qual #2 here. Please indicate authorship!
Contents
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
Problem 4
Problem 5
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 $. If $ n=0 $, then $ f $ is constant, again a contradiction of one-to-one; so $ n=1 $, and $ f= ax+b $ for some $ a \neq 0,b $ in $ \mathbb C $.
