MATH3401
Complex Analysis
Lecture 9
Branches
Branch: interval of the form
\(
\alpha \leq \hat{\theta} < \alpha + 2 \pi
\),
or
\(\alpha < \hat{\theta} \leq \alpha + 2 \pi\),
for some \(\alpha \in \R.\)
We can define a single-valued \(\,\arg\) with values in that interval and hence a single-valued \(\,\log\), and hence a single-valued branch of, e. g., \(z^{1/2}\).
Branch cut
\(\left\{ z: \arg(z) = \alpha \right\} \cup \left\{ 0 \right\}\) subset of \(\C\).
E. g. PV\(\left(z^{1/2}\right)\): \[ \begin{align*} & z\mapsto |z|^{1/2}\exp\left(\frac{i\Arg (z)}{2}\right) \\ &re^{i \theta} \mapsto \sqrt{r}\exp \left(\frac{i\theta}{2}\right), -\pi < \theta \leq \pi. \end{align*} \]
Branch is \(-\pi < \theta \leq \pi\), and the branch cut is the negative real axis \(\cup \left\{ 0 \right\}\).
Branch cut
Trigonometric functions
For \(x\in \R\) \[ \begin{align*} e^{ix} &= \cos x + i \sin x \quad (1)\\ e^{-ix} &= \cos x - i \sin x \quad (2) \end{align*} \]
\((1) + (2)\) implies \[ \cos x = \frac{e^{ix}+ e^{-ix}}{2} \quad (3)' \]
\((1) - (2)\) implies \[ \sin x = \frac{e^{ix} - e^{-ix}}{2i} \quad (4)' \]
Trigonometric functions
Use (3)' and (4)' to define \(\cos\) and \(\sin\) on \(\C\), i. e.: \[ \begin{align*} \cos z &= \frac{e^{iz}+ e^{-iz}}{2}\\ \sin z &= \frac{e^{iz}- e^{-iz}}{2i}. \end{align*} \]
Trigonometric functions
\[ \begin{align*} \cos z &= \cos(-z) \quad (3)\\ \sin z &= -\sin(-z) \quad (4) \end{align*} \]
\[ \begin{align*} \cos (z + \xi) &= \cos z \cos \xi - \sin z \sin \xi \quad (5)\\ \sin (z + \xi) &= \sin z \cos \xi + \cos z \sin \xi \quad (6) \end{align*} \]
\[ \begin{align*} \sin^2 z + \cos^2 z &= 1 \quad (7) \end{align*} \]
\[ \begin{align*} \sin \left(z+\frac{\pi}{2}\right) &= \cos z \quad (8)\\ \sin \left(z-\frac{\pi}{2}\right) &= -\cos z \quad (9) \end{align*} \]
Proof: Use properties of \(\exp\) 📝.
Hyperbolic functions
Recall hyperbolic functions on \(\R\):
|
\(y = \sinh x = \dfrac{e^x-e^{-x}}{2}\) 
|
\(y = \cosh x = \dfrac{e^x+e^{-x}}{2}\) 
|
|---|
Hyperbolic functions
On \(\C\), define \[ \sinh z = \frac{e^z-e^{-z}}{2}\quad \text{&} \quad \cosh z = \frac{e^z+e^{-z}}{2}. \]
So \[ \begin{align*} \sin (iy) &= \frac{e^{-y}-e^{y}}{2i} = \frac{i(e^y - e^{-y})}{2} \\ &= i \sinh y. \qquad\qquad \quad (10) \end{align*} \]
\[ \begin{align*} \cos (iy) &= \frac{e^{-y}+e^{y}}{2} \\ &= \cosh y. \qquad\qquad \quad (11) \end{align*} \]
Hyperbolic functions
Take \(z= x\) and \(\xi = iy\) in (5), (6) \[ \sin (x+iy) \overset{{\tiny(6)}}{=} \sin x \cos (iy) + \cos x \sin (iy), \quad \quad (12) \]
\(\qquad \qquad \qquad \overset{{\tiny(12)(11)}}{=} \sin x \cosh y + i \cos x \sinh y\).
Similarly 📝 \[ \cos (x+iy) = \cos x \cosh y - i \sin x \sinh y. \quad \quad (13) \]
Hyperbolic functions
(12) and (13) implies \[ \begin{align*} \sin (z + 2\pi) &= \sin z, \quad \quad (14)\\ \cos (x+2\pi ) &= \cos z. \quad \quad (15) \end{align*} \]
Note that \(\cosh^2 z = 1 + \sinh^2 z\) 📝.
Using this fact we have
\( \qquad
|\sin z |^2 \overset{{\tiny(12)}}{=} \sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y
\)
\( \qquad \qquad \quad = \sin^2 x \left(1+\sinh^2 y\right) + \left(1- \sin^2 x\right) \sinh^2y \)
\( \qquad \qquad \quad = \sin^2 x + \sinh^2 y. \)
Similarly \( |\cos z |^2 = \cos^2 x + \sinh^2 y. \)
Bounded functions
Define \(\,f: \Omega \ra \C\) is bounded if exists \(M\) such that \[ |f(z)|\leq M \quad \forall z\in \Omega. \]
Bounded functions
Bounded functions
The functions \(\sin \) & \(\cos \) are unbounded on \(\C\).