Lecture 16
\(\ds \frac{d}{dz}\log z\): Consider \(|z|>0\) and recall
\[ \begin{align} \log z & = \ln |z| + i \,\arg(z) \\ & = \ln r + i \theta. \end{align} \]
\( \implies u = \ln r, v = \theta\)
\( \implies u_r = \ds \frac{1}{r}, u_{\theta} = 0 , v_r =0, v_{\theta} = 1.\)
C/R in polar coordinates: \[ \begin{align} ru_r & = v_{\theta} \\ u_{\theta} & = -ru_r \end{align} \]
Satisfies sufficient conditions for complex differentiability on any subset of \(\C_*\), such that \(\alpha \lt \theta \lt \alpha + 2 \pi\), with \(\alpha\) fixed in \(\R\).
So if we specify a branch, we have that \(\log\) is differentiable and expression (6) in Lecture 15 \( \implies \)
\( \ds \frac{d}{dz}\log z \) \( = e^{-i\theta} \left( u_r + i v_r \right)\) \( = e^{-i\theta}\ds \frac{1}{r}\) \( = \ds \frac{1}{re^{i\theta}}\) \( = \ds \frac{1}{z}\)
E. g. \( \ds \frac{d}{dz}\Log\, z \) \( \ds = \frac{1}{z} \)
for \( -\pi \lt \Arg \, z \lt \pi \), \(|z|> 0\).
For \(f(z)= z^c\), with \(c\in \C_*\) fixed, defined on \(\C_*\):
\[f(z) = \exp\left( c \log z \right) \]
\(f'(z) = \exp\left( c \log z \right) \cdot \ds \frac{c}{z} \) \((*)\)
\(\qquad = z^c \cdot \ds \frac{c}{z} \) \( = c z^{c-1} \) \((**)\)
\((**)\) is valid on any domain of the form \[\left\{ z: |z|>0, \alpha \lt \arg\, z \lt \alpha + 2\pi\right\}\]
(due to needing to pick a branch of \(\log z\) in \((*)\)).
📝Remark: Try for \(g(z) = c^z\)
\(\Omega \subseteq \R^n\)
\(*\) \(\mathcal C\left( \Omega \right)\) \( = \mathcal C^0\left( \Omega \right)\)
\(\quad\quad\;\, = \left\{ \text{continuous functions}: \Omega \ra \R \right\}\).
\(*\) \( \mathcal C^k\left( \Omega \right)\) \( = \left\{ \text{functions } f:\Omega \ra \R:\right. \) \(\qquad \quad\left. f \text{ and all its derivatives/partial derivatives } \right.\) \(\qquad \quad\left. \text{of order } 0,1,\ldots, k \text{ exist and are continuous} \right\}\).
Note: \(0\)th order derivative of \(f\) is \(f\).
\(*\) \( \mathcal C^{\infty}\left( \Omega \right)\) \( = \left\{ \text{functions } f:\Omega \ra \R:\right. \) \(\qquad \quad\left. f \text{ and all its derivatives/partial derivatives } \right.\) \(\qquad \quad\left. \text{of all orders exist and are continuous} \right\}\).
a.k.a. Smooth functions
\(*\) \( f\in \mathcal C^{\omega} \left( \Omega \right)\) says at every \(x_0\in \Omega\):
Note: (i) \(\Rightarrow f\in C^{\infty}(\Omega)\) , (i) \(\nRightarrow \) (ii) in \(\R^{n}\).
Consider \[ f(x) = \left\{ \begin{array}{lr} e^{-1/x^2} & x > 0\\ 0 & x \le 0 \end{array} \right. \]
📝Check: \(f^{(n)}(x)\) exists for every \(x\neq 0\) , and \(f^{(n)}(0) = 0\) \(\,\forall n\in \N\), and \(f^{(n)}\) is continuous on \(\R\). Then Taylor series for \(f\) at \(0\) is \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \equiv 0. \]
\(f\in \mathcal C^{\infty}(\R), \quad f\notin \mathcal C^{\omega}(\R)\).
\(\quad\)
\(\mathcal C^{\omega} \) \(\subsetneq \mathcal C^{\infty}\) \(\subsetneq \ldots \subsetneq \mathcal C^{1337} \subsetneq \mathcal C^{1336}\)
\(\subsetneq \ldots \subsetneq \mathcal C^{1} \subsetneq \mathcal C^{0}\).
\(\quad\)