Lecture 10
\((*)\) \(\;f(z)=\pi\) 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 \(\C\).
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: \[ \begin{align*} \sinh: & \text{ only } n\pi i, \quad n\in \Z.\\ \cosh: & \text{ only } \left(n+\frac{1}{2}\right)\pi i, \quad n\in \Z. \end{align*} \]
Set \(w = \sin ^{-1}z\), then
\(z = \sin w\) \(= \dfrac{e^{iw}- e^{-iw}}{2i}\cdot \dfrac{e^{iw}}{e^{iw}}\) \( \;\; = \dfrac{e^{2w}-1}{2i e^{iw}} \)
\(\implies 2iz e^{iw} = e^{2iw} - 1 \)
\(\implies \left( e^{iw} \right)^2- 2i z\left(e^{iw}\right) -1 = 0. \)
\(\implies e^{iw} = \dfrac{2iz + \left( -4z^2+4\right)^{1/2}}{2}\)
\( \qquad \qquad = iz + \left( 1-z^2\right)^{1/2}\)
\( \implies iw = \log \left( iz + \left( 1-z^2\right)^{1/2} \right)\)
\(\implies w = \sin^{-1} z = -i \log \left( iz + \left( 1-z^2\right)^{1/2} \right) \)
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 \[ \sin^{-1}(-i) = \left\{ \begin{align*} & -i\left( \ln\left(1+\sqrt{2} \right) + 2n\pi i \right), n\in\Z;\\ & -i\left( \ln\left(\sqrt{2} - 1 \right) + (2n+1)\pi i \right), n \in \Z. \end{align*} \right. \]
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 \[\{z: 0 \lt |z-z_0| < \varepsilon\}.\]
\( B_{\varepsilon}(z_0) = \{z: |z-z_0| < \varepsilon\}\)
\(|z-z_0| = \left| \left( x+iy\right) - \left( x_0+iy_0\right) \right| \) \(= \left| \left( x-x_0\right) + \left( y +y_0\right)i \right| \)
\(\qquad \quad = \sqrt{ \left( x-x_0\right)^2 + \left( y -y_0\right)^2 } \)
\(\qquad \quad = \left|\left| \left( x,y\right) - \left( x_0 ,y_0\right) \right|\right|_{\R^2} \)
\(\qquad \quad = d\left( \left( x,y\right) , \left( x_0 ,y_0\right) \right)_{\R} \)
\(\qquad \quad = d\left( z, z_0 \right)_{\C} \).
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. \(\C \setminus \Omega\).