# MATH3401

### Complex Analysis

Lecture 10

#### Bounded & unbounded functions

$$(*)$$ $$\;f(z)=\pi$$ is bounded on $$\C$$.

$$(*)$$ $$\;f(z)=1/z$$ is unbounded on $$\C_{*}$$,
$$(*)$$ $$\;f(z)=1/z$$ is unbounded on $$\{z: 0 \lt |z| \leq 1\}$$,
$$(*)$$ $$\;f(z)=1/z$$ is bounded on $$\{z: |z| \geq 1\}$$,
$$(*)$$ $$\;f(z)=1/z$$ is bounded on $$\{z: |z| = 1\}$$.

$$(*)$$ $$\;f(z)=z$$ is unbounded on $$\C$$, bounded on any bounded subset of $$\C$$.

$$(*)$$ $$\;f(z)=\dfrac{1}{1+|z|}$$ is bounded on $$\C$$.

#### Zeros of functions

Definition: A zero of a function $$f$$ is a value of $$z$$ such that $$f(z)=0$$.

E. g. Zeros of $$\sin$$: $(12)\implies n\pi + 0 i, \quad n\in \Z.$

E. g. Zeros of $$\cos$$: $(13)\implies \text{ only } \left(n+\frac{1}{2}\right)\pi , \quad n\in \Z.$

Similarly zeros of: \begin{align*} \sinh: & \text{ only } n\pi i, \quad n\in \Z.\\ \cosh: & \text{ only } \left(n+\frac{1}{2}\right)\pi i, \quad n\in \Z. \end{align*}

##### Inverse trigonometric functions
###### $$\sin^{-1}$$, $$\cos^{-1}$$

Set $$w = \sin ^{-1}z$$, then

$$z = \sin w$$ $$= \dfrac{e^{iw}- e^{-iw}}{2i}\cdot \dfrac{e^{iw}}{e^{iw}}$$ $$\;\; = \dfrac{e^{2w}-1}{2i e^{iw}}$$

$$\implies 2iz e^{iw} = e^{2iw} - 1$$

$$\implies \left( e^{iw} \right)^2- 2i z\left(e^{iw}\right) -1 = 0.$$

##### Inverse trigonometric functions
###### $$\sin^{-1}$$, $$\cos^{-1}$$

$$\implies e^{iw} = \dfrac{2iz + \left( -4z^2+4\right)^{1/2}}{2}$$

$$\qquad \qquad = iz + \left( 1-z^2\right)^{1/2}$$

$$\implies iw = \log \left( iz + \left( 1-z^2\right)^{1/2} \right)$$

$$\implies w = \sin^{-1} z = -i \log \left( iz + \left( 1-z^2\right)^{1/2} \right)$$

##### Inverse trigonometric functions
###### $$\sin^{-1}$$, $$\cos^{-1}$$

Example: $$\sin^{-1}(-i)$$ $$= -i\log\left(1+2^{1/2}\right)$$ $$= -i\log\left(1\pm \sqrt{2}\right)$$.

Now, $$\log\left( 1+\sqrt{2} \right) = \ln\left( 1+\sqrt{2} \right) + 2n\pi i$$ with $$n \in \Z$$,

and $$\log\left(1 -\sqrt{2} \right) = \ln\left(\sqrt{2} - 1 \right) + (2n+1)\pi i$$ with $$n \in \Z$$.

So \sin^{-1}(-i) = \left\{ \begin{align*} & -i\left( \ln\left(1+\sqrt{2} \right) + 2n\pi i \right), n\in\Z;\\ & -i\left( \ln\left(\sqrt{2} - 1 \right) + (2n+1)\pi i \right), n \in \Z. \end{align*} \right.

#### Topology

Given $$z_0\in \C$$ and $$\varepsilon > 0$$, $$B_{\varepsilon}(z_0)$$ denotes the (open) ball of radius $$\varepsilon$$ about $$z_0$$, a.k.a $$\varepsilon$$-neighbourhood of $$z_0$$, given by $\{z: |z-z_0| < \varepsilon\}.$

$$*\;$$ $$\overline{B}_{\varepsilon}(z_0) =$$ closed ball of radius $$\varepsilon$$ about $$z_0$$, a.k.a closed $$\varepsilon$$-neighbourhood of $$z_0$$, given by $\{z: |z-z_0| \leq \varepsilon\}.$

$$*\;$$ A deleted $$epsilon$$ neighbourhood of $$z_0$$ is given by $\{z: 0 \lt |z-z_0| < \varepsilon\}.$

#### Topology

$$B_{\varepsilon}(z_0) = \{z: |z-z_0| < \varepsilon\}$$

$$|z-z_0| = \left| \left( x+iy\right) - \left( x_0+iy_0\right) \right|$$ $$= \left| \left( x-x_0\right) + \left( y +y_0\right)i \right|$$

$$\qquad \quad = \sqrt{ \left( x-x_0\right)^2 + \left( y -y_0\right)^2 }$$

$$\qquad \quad = \left|\left| \left( x,y\right) - \left( x_0 ,y_0\right) \right|\right|_{\R^2}$$

$$\qquad \quad = d\left( \left( x,y\right) , \left( x_0 ,y_0\right) \right)_{\R}$$

$$\qquad \quad = d\left( z, z_0 \right)_{\C}$$.

#### Topology

Take $$\Omega \subseteq \C$$.

$$*\;$$ $$z\in \C$$ is an interior point of $$\Omega$$ if $$\exists \varepsilon > 0$$ such that $$B_{\varepsilon}(z)\subset \Omega$$

(note $$\Rightarrow B_{\varepsilon^{\prime}}(z)\subset \Omega\; \forall \varepsilon^{\prime}$$, $$0 < \varepsilon^{\prime} < \varepsilon$$).

$$*\;$$ $$z\in \C$$ is an exterior point of $$\Omega$$ if $$\exists \varepsilon > 0$$ such that $B_{\varepsilon}(z)\cap \Omega \neq \emptyset$

$$*\;$$ $$z\in \C$$ is a boundary point of $$\Omega$$, $$z\in\partial \Omega$$, if $$\forall \varepsilon>0$$:
$B_{\varepsilon}(z)\cap \Omega \neq \emptyset \text{ and } B_{\varepsilon}(z)\cap \Omega^c \neq \emptyset.$

$$\Omega^c$$ is the complement of $$\Omega$$, i. e. $$\C \setminus \Omega$$.