4.1 KiB
Math416 Lecture 16
Answer checking for exam
Q1
Cauchy riemann equations:
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\quad\text{and}\quad\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
Liouville's Theorem:
Any non-constant entire function is unbounded.
So \cos(z) is unbounded in \mathbb{C}.
\text{Log}(-e^2) = \ln|-e^2| + i\arg(-e^2) = -2 + \pi i
At any point z_0\in \mathbb{C}\setminus\{0\}, there is an open set z_0\in U\subset \mathbb{C} and a branch of logarithm defined on U.
Q2
Power series expansion
Q3
limit superior
Q4
Bound integral
Q5
f_n converges pointwise to f on U if \forall z\in U, \forall \epsilon > 0, \exists N s.t. \forall n\geq N, |f_n(z)-f(z)| < \epsilon.
f_n converges uniformly to f on U if \forall \epsilon > 0, \exists N s.t. \forall n\geq N, \forall z\in U, |f_n(z)-f(z)| < \epsilon.
Show for |z|<1, f_n(z)=z^n converges pointwise to 0 but not uniformly to 0.
(a) pointwise convergence:
|z^n| = |z|^n < \epsilon if n > \frac{\ln\epsilon}{\ln|z|}.
(b) uniform convergence:
No matter how small \epsilon is, there is always a z s.t. |z^n| > \epsilon for all n.
Continue from last lecture
Schwarz's Lemma
Let f be an holomorphic function that maps the unit disk D(0,1) to itself and f(0)=0. Then |f(z)|\leq |z| for all z\in D(0,1)
Schwarz-Pick's Lemma
(see exercise 7.17.2)
Let f be an holomorphic function that maps the unit disk D(0,1) to itself. Then \forall z,w\in D(0,1),
\left|\frac{f(z)-f(w)}{1-\overline{f(w)}f(z)}\right|\leq \left|\frac{z-w}{1-\overline{w}z}\right|
Recall the Möbius map
\phi_\alpha(z) = \frac{z-\alpha}{1-\overline{\alpha}z}is a homeomorphism of the unit disk.
So we can use the Möbius to restate the Schwarz-Pick's Lemma as:
|\phi_{f(w)}(f(z))|\leq |\phi_w(z)|
Suppose we defined g=\phi_{f(w)}\circ f\circ \phi_{-w}, then g is a holomorphic function that maps the unit disk to itself and g(0)=0.
By Schwarz's Lemma, let z\in D(0,1), |g(z)|\leq |z|.
|\phi_{f(w)}(f(\phi_{-w}(z)))|\leq |z|
Let \zeta=\phi_{-w}(z), then \zeta=\frac{z+w}{1+\overline{w}z}\in D(0,1), so |\zeta|=\phi_w(z).
Extension of Schwarz-Pick's Lemma in hyperbolic metric
Suppose we defined the distance on \mathbb{C} as d(z,w)=|\frac{z-w}{1-\overline{w}z}|.
We claim that this is a metric on \mathbb{C}. \forall z,w,v\in \mathbb{C}:
(a) d(z,w)=0 if and only if z=w and d(z,w)> 0 otherwise.
(b) d(z,w)=d(w,z).
(c) d(z,w)\leq d(z,v)+d(v,w).
We call this metric the Pseudo hyperbolic metric.
Hyperbolic metric:
\text{Hypdist}(z,w)=\tanh^{-1}(d(z,w))Where
d(z,w)=|\frac{z-w}{1-\overline{w}z}|
So we can restate the Schwarz-Pick's Lemma as:
d(f(z),f(w))\leq d(z,w)
And in hyperbolic metric, it becomes:
\text{Hypdist}(f(z),f(w))\leq \text{Hypdist}(z,w)
Suppose the equality holds for Schwarz-Pick's Lemma, then |g(z)|=\tau z where |\tau|=1.
Computation ignored here.
Then f is a Möbius map that is automorphism of the unit disk.
Existence of harmonic conjugate
Suppose f=u+iv is holomorphic on a domain U\subset \mathbb{C}. Then u=\text{Re}(f) is harmonic on U. That is \Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.
Theorem 7.18
Let u be a real harmonic function on a convex domain G\subset \mathbb{C}. Then there exists g\in O(G) such that \text{Re}(g)=u. Moreover, g is unique up to an additive imaginary constant.
Proof:
Existence next time.
Uniqueness:
Suppose g,h\in O(G) s.t. \text{Re}(g)=\text{Re}(h)=u.
\text{Re}g=u=\text{Re}h on G.
If we can show that (g-h)'=0 on G, then we win.
Let g=u+iv, h=u+iw.
By the Cauchy-Riemann equations,
\begin{aligned}
\frac{\partial}{\partial x}(g-h)&=\frac{\partial}{\partial x}i(v-w)\\
&=i\left(\frac{\partial u}{\partial y}-\frac{\partial u}{\partial y}\right)\\
&=0
\end{aligned}
Suppose G=\mathbb{C}\setminus\{0\}, then u=\ln|z|=\frac{1}{2}\ln(x^2+y^2), which is harmonic.
Continue next time.