Lecture 15
in polar coordinates
\(z = x+iy=r e^{i\theta }\), \(x = r\cos \theta\), \(y = r \sin \theta\).
Chain rule implies
\( u_r = u_x \cos \theta + u_y \sin \theta \quad \quad (1)\)
\( u_{\theta} = -u_x r \sin \theta + u_y r \cos \theta \quad (2)\)
\( v_r = v_x \cos \theta + v_y \sin \theta \quad \quad (3)\)
\( v_{\theta} = -v_x r \sin \theta + v_y r \cos \theta \quad (4)\)
in polar coordinates
Combine with C/R \(\implies\) C/R in polar coordinates:
\[ \begin{align} ru_r & = v_{\theta} \\ u_{\theta} & = -ru_r \end{align} \]
in polar coordinates
Useful: (recall (⚛️), Lecture 14)
If \(f'\) exists then \[ \begin{align} f' & = u_x + i v_x \qquad (5), \end{align} \] and similarly 📝 \[ \begin{align} f'(z) & = e^{-i\theta}\left( u_r + i v_r\right) \qquad (6) \end{align} \]
Formally: change of variables \( \left(x,y\right) \mapsto \left(z,\overline{z}\right) \)
\[z= x+iy, \quad \overline{z}= x-iy\]
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\(\ds \frac{\partial f}{\partial x} \) \(= \ds \frac{\partial f}{\partial z} \frac{\partial z}{\partial x} + \frac{\partial f}{\partial \overline{z}} \frac{\partial \overline{z}}{\partial x} \) \(\quad\; = \ds \frac{\partial f}{\partial z} + \frac{\partial f}{\partial \overline{z}} \quad (7) \) |
\(\ds \frac{\partial f}{\partial y} \) \(= \ds \frac{\partial f}{\partial z} \frac{\partial z}{\partial y} + \frac{\partial f}{\partial \overline{z}} \frac{\partial \overline{z}}{\partial y} \) \(\quad\; = \ds i \frac{\partial f}{\partial z} - i \frac{\partial f}{\partial \overline{z}} \quad (8) \) |
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(7)\(-i\)(8) \(\implies\) \( \ds \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\) \(= 2 \ds \frac{\partial f}{\partial z} \) \( \implies \ds \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) \)
Finally (7)\(+i\)(8) \(\implies\) \( \ds \frac{\partial}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) \)
\( \ds \frac{\partial}{\partial z}, \frac{\partial}{\partial \overline{z}}\) are called the Wirtinger operators.
Example
\(f(z) = z^n = (x+iy)^n \) with \(n \in \Z\).
\(\ds \frac{\partial f}{\partial z} \) \( = \ds \frac12 \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right) (x+iy)^n \) \(= \ds \frac12 n(x+iy)^{n-1}\left(1-i^2\right) \)
\( \quad \;= \ds n(x+iy)^{n-1} \) \( = nz^{n-1} \) \( = f'(z) \).
\(\ds \frac{\partial f}{\partial \overline{z}} =0\).
Relationship with C/R
For \(f = u+iv \) complex differentiable:
\(\ds \frac{1}{2} \frac{\partial f}{\partial x} \) \( = \ds \frac12 \left( u_x + iv_x \right) \) \(= \ds \frac12 \left( v_y- iu_y \right) \)
\( \qquad \;= \ds - \frac{i}{2} \left( u_y+ iv_y \right) \) \( = \ds - \frac{i}{2} \frac{\partial f}{\partial y} \).
So C/R \(\iff \ds \frac{\partial f}{\partial \overline{z}} =0\).
Relationship with C/R
Note also from (5):
\(\ds \frac{d f}{d z} \) \( = u_x + iv_x \) \(= \ds \frac{\partial f}{\partial x}\) \(= \ds - i \frac{\partial f}{\partial y} \) \(= \ds \frac{i}{2} \left( \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y} \right) \) \( = \ds \frac{\partial f}{\partial z} \).
Application 1
Find \(f'(z)\) for \(f(z) = \exp z\).
Check sufficient conditions for \(f'\) to exist:
a) \(f= u+iv \) \( = e^{x+iy} \) \( = e^x \left(\cos y +i \sin y\right) \) defined on \(\C\).
\( u = e^x\cos y, v = e^{x}\sin y \). \(u_x, u_y, v_x, v_y \) defined and continuous on \(\C\).
b) Check C/R: \(u_x=v_y\) & \(u_y = -v_x \) or \( \ds \frac{\partial f}{\partial \overline{z}}=0. \)
Application 2
When is \(g(z) = |z|^2\) differentiable?
Note: \(g(z)=z\overline{z} = x^2+y^2\).
C/R: \(\ds \frac{\partial g}{\partial \overline{z}}= 0 \) \( \implies z = 0 \)
So \(g\) cannot be differentiable for \(z\neq 0\) because C/R is necessary for differentiability.
Application 2
At \(z=0\) check sufficient conditions:
\(u,v\) are defined on \(\C\). \(u_x,u_y,v_x,v_y\) are defined and continuous on \(\C\) (a neighbourhood of \(0\)).
C/R hold. So \(g'(0)=0\).
📝Exercise: Go through Application 1 for \(z\mapsto 1/z\) on \(\C_*\)
Definition: \(f:\Omega \ra \C\) is analytic at \(z_0\) if \(f\) is differentiable on a neighbourhood of \(z_0\).
A function is singular at \(z_0\) if it is NOT analytic at \(z_0\), but is analytic at some point in every neighbouhood of \(z_0\).
E. g. \(f(z)=\ds \frac{1}{z}\) is analytic on \(\C_*\) and singular at \(0\) on \(B_{\varepsilon}(0)\), \(f\) is analytic on \(B_{\varepsilon'}(z_0)\), \(z_0\in B_{\varepsilon}(0)\), \(\varepsilon'\lt |z_0|\).
A function is entire if it is analytic on all of \(\C\).
E. g. Polynomials, \(\sin,\cos, \exp, \cosh, \sinh\).
Note: If a function is differentiable at precisely one point,
it is not analytic there, or anywhere. E. g. \(|z|^2\).