# Complex-Differentiability

## By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

### November 28, 2017

# Contents

# Exercises

## Antiholomorphic Functions

A function \(f:\Omega \to \mathbb{C}\) is *antiholomorphic* if its complex conjugate \(\overline{f}\) is holomorphic.

### Questions

Is the complex conjugate function \(c:z \in \mathbb{C} \mapsto \overline{z}\) real-linear? complex-linear? Is it real-differentiable? holomorphic? antiholomorphic?

Show that any antiholomorphic function \(f\) is real-differentiable. Relate the differential of such a function and the differential of its complex conjugate.

Find the variant of the Cauchy-Riemann equation applicable to antiholomorphic functions.

What property has the composition of two antiholomorphic functions?

Let \(f: \Omega \mapsto \mathbb{C}\) be a holomorphic function; show that the function \[g: z \in \overline{\Omega} \mapsto \overline{f(\overline{z})}\] is holomorphic and compute its derivative.

### Answers

For any \(\lambda \in \mathbb{R}\) and \(w, z \in \mathbb{C}\), we have \[ \overline{w + z} = \overline{w} + \overline{z} \; \wedge \; \overline{\lambda z} = \lambda \overline{z}, \] therefore the function \(c\) is real-linear. However, it is not complex-linear: for example \(\overline{i} = - i \neq i \times \overline{1}.\) The function \(c\) is real-linear and continuous, hence it is real-differentiable and for any \(z \in \mathbb{C}\), \(dc_z = c\). This differential is not complex-linear, therefore the function is not complex-differentiable (or holomorphic). On the other hand, \(\overline{c(z)} = z\), therefore it is antiholomorphic.

If the function \(\overline{f}:\Omega \to \mathbb{R}\) is holomorphic, it is real-differentiable everywhere on its domain of definition. Hence \(f = c \circ \overline{f}\) is real-differentiable as the composition of real-differentiable functions and \[df_z = d(c\circ \overline{f})_z = c \circ d\overline{f}_z.\]

The complex-valued Cauchy-Riemann equation for \(\overline{f}\) is \[ \frac{\partial \overline{f}}{\partial x}(z) = \frac{1}{i}\frac{\partial \overline{f}}{\partial y}(z), \; \mbox{ or } d \overline{f}_z(i) = i \times d \overline{f}_z(1) \] On the other hand, we have \[ \frac{\partial \overline{f}}{\partial x}(z) = d (c \circ f)_z (1) = (c \circ df_z)(1) = \overline{\frac{\partial f}{\partial x}} \] and \[ \frac{\partial \overline{f}}{\partial y}(z) = d (c \circ f)_z (i) = (c \circ df_z)(i) = \overline{\frac{\partial f}{\partial y}}, \] hence we can rewrite the Cauchy-Riemann equation for \(\overline{f}\) as \[ \frac{\partial f}{\partial x}(z) = -\frac{1}{i}\frac{\partial f}{\partial y}(z), \; \mbox{ or } d f_z(i) = -i \times d f_z(1). \]

The composition of antiholomorphic functions is holomorphic. Indeed, if \(f\) and \(g\) are antiholomorphic, they are real-differentiable; their composition – assuming that it is defined – is \(f \circ g = (c \circ \overline{f}) \circ (c \circ \overline{g})\); it satisfies \[ d(f \circ g)_z =d(c \circ \overline{f})_{c(\overline{g}(z))} \circ d(c \circ \overline{g})_z = c \circ d\overline{f}_{c(\overline{g}(z))} \circ c \circ d\overline{g}_z. \] Since for any \(h, w \in \mathbb{C}\), \[c(h w) = \overline{h} c(w), \;d\overline{g}_z(h w) = h d\overline{g}(w), \; d\overline{f}_{c(\overline{g}(z))}(h w) = h d\overline{f}_{c(\overline{g}(z))}(w),\] we have \[ d(f \circ g)_z(h) = h d(f \circ g)_z(1). \] The differential of \(f \circ g\) is complex-linear: the function is holomorphic.

If the point \(w\) belongs to \(\overline{\Omega}\), then \(w = \overline{z}\) for some \(z \in \Omega\), thus the complex number \(\overline{f(\overline{z})}\) is defined. Additionally, the set \[\overline{\Omega} = \{\overline{z} \;|\; z \in \Omega\} = \{w \in \mathbb{C}\; | \; \overline{w} \in \Omega\}\] is an open set, as the inverse image of the open set \(\Omega\) by the continous function \(c\). The function \(g\) satisfies \[ g = c \circ f \circ c = (c \circ f) \circ c \] which is holomorphic as the composition of two antiholomorphic functions. We have \[ dg_z(h) = (c \circ df_{c(z)} \circ c) (h) = \overline{df_{\overline{z}}(\overline{h})} = \overline{f'(\overline{z}) \overline{h}} = \overline{f'(\overline{z})} h, \] hence \(g'(z) = \overline{f'(\overline{z})}\).

## Principal Value of the Logarithm

According to the definition of \(\log\), for any \(x+iy \in \mathbb{C} \setminus \mathbb{R}_-\), \[
\log (x + i y)
=
\ln \sqrt{x^2 + y^2} + i \arg (x+iy),
\] where \(\arg: \mathbb{C} \setminus \mathbb{R}_- \to \mathbb{C}\) is *the principal value of the argument*: \[
\forall\, z \in \mathbb{C} \setminus \mathbb{R}_-, \;
\arg z \in \left]-\pi,\pi\right[ \; \wedge \; e^{i \arg z} = \frac{z}{|z|}.
\]

### Questions

Show that \[ \arg (x + i y) = \left| \begin{array}{cl} \arctan y/x & \mbox{if } x > 0, \\ +\pi/2 - \arctan x/y & \mbox{if } y > 0, \\ -\pi/2 - \arctan x/y & \mbox{if } y < 0. \\ \end{array} \right. \]

Show that the function \(\log\) is holomorphic and compute its derivative.

### Answers

Note that the definition of \(\arg\) is non-ambiguous: for any nonzero real number \(\epsilon\), \[ \arctan \epsilon + \arctan 1/\epsilon = \mathrm{sgn}(\epsilon) \times \pi/2, \] so if \(x+iy\) belongs to two of the half-planes \(x>0\), \(y<0\) and \(y>0\), the two relevant expressions which may define \(\arg(x+iy)\) are equal.

As \(\arctan(\mathbb{R}) = \left]-\pi/2, \pi/2\right[\), the three expressions that define \(\arg\) have values in \(\left]-\pi,\pi\right[\). Then, if for example \(x>0\), with \(\theta=\arg(x+iy)\), we have \[ \frac{\sin\theta}{\cos\theta}=\tan \theta = \tan (\arctan y/x) = \frac{y}{x}. \] Since \(\cos \theta > 0\) and \(x>0\), there is a \(\lambda > 0\) such that \[ x + i y = \lambda (\sin \theta + i \cos \theta) = \lambda e^{i\theta}; \] this equation yields \[e^{i\arg (x+ i y)} = \frac{x+iy}{|x+iy|}.\] The proof for the half-planes \(y>0\) and \(y<0\) is similar.

The functions \(\arg\), \(\ln\) and therefore \(\log\) are continuously real-differentiable. If \(x > 0\), for example, we have \[ \frac{\partial \arg(x+iy)}{\partial x} = \frac{1}{1 + (y/x)^2} \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2}. \] and \[ \frac{\partial \arg(x+iy)}{\partial y} = \frac{1}{1 + (y/x)^2} \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2}. \] On the other hand, \[ \frac{\partial}{\partial x} \ln \sqrt{x^2 + y^2} = \frac{1}{\sqrt{x^2 + y^2}} \frac{2x}{\sqrt{x^2 + y^2}} = \frac{x}{x^2 + y^2} \] and \[ \frac{\partial}{\partial y} \ln \sqrt{x^2 + y^2} = \frac{1}{\sqrt{x^2 + y^2}} \frac{2y}{\sqrt{x^2 + y^2}} = \frac{y}{x^2 + y^2} \] Finally \[ \frac{\partial \log}{\partial x}(x+iy) = \frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2} = \frac{1}{x+iy} \] and \[ \frac{\partial \log}{\partial y}(x+iy) = \frac{y}{x^2+y^2} + i \frac{x}{x^2+y^2} =\frac{1}{i} \frac{1}{x+iy}. \] The computations for \(y>0\) and \(y<0\) are similar. Conclusion: the function \(\log\) is complex-differentiable and \(\log'(z) = 1/z\).

## Conformal Mappings

### Questions

A \(\mathbb{R}\)-linear mapping \(L: \mathbb{C} \to \mathbb{C}\) is*angle-preserving*if \(L\) is invertible and \[ \forall \, \theta \in \mathbb{R}, \; \exists \, \alpha_{\theta} > 0, \; L(e^{i\theta}) = \alpha_{\theta} \times e^{i\theta} L(1). \] A \(\mathbb{R}\)-differentiable function \(f: \Omega \to \mathbb{C}\)

*(locally) angle-preserving*– or

*conformal*– if its differential is angle-preserving everywhere.

Show that an invertible \(\mathbb{R}\)-linear mapping \(L: \mathbb{C} \to \mathbb{C}\) is angle-preserving if and only if it is \(\mathbb{C}\)-linear.

Identify the class of conformal mappings defined on \(\Omega\).

### Answers

If \(L\) is \(\mathbb{C}\)-linear, then for any \(\theta \in \mathbb{R}\), \(L(e^{i\theta}) = e^{i\theta} L(1)\), hence it is angle-preserving. Reciprocally, if \(L\) is angle-preserving, we have on one hand \[ L(e^{i \theta}) = \alpha_{\theta} \times e^{i\theta} L(1) \] and on the other hand, as \(e^{i\theta}= \cos \theta + i \sin \theta\), \[ L(e^{i\theta}) = \cos \theta \times L(1) + \sin \theta \times L(i) = (\cos \theta + \sin \theta \times \alpha_{\frac{\pi}{2}} i) L(1). \] We know that \(L(1) \neq 0\), hence these equations provide \[ \alpha_{\theta} = \cos \theta e^{-i\theta} + \sin \theta e^{-i \theta}\times \alpha_{\frac{\pi}{2}} i = \frac{1 + \alpha_{\frac{\pi}{2}}}{2} + \frac{1 - \alpha_{\frac{\pi}{2}}}{2} e^{-i2\theta}. \] As \(\alpha_{\theta}\) is real-valued, \(\alpha_{\frac{\pi}{2}} = 1\). Consequently \(\alpha_{\theta} = 1\) and \(L\) is \(\mathbb{C}\)-linear.

A mapping \(f:\Omega \to \mathbb{C}\) is conformal if it is \(\mathbb{R}\)-differentiable, \(df_z\) is invertible everywhere and is \(\mathbb{C}\)-linear: this is exactly the class of holomorphic mappings \(f\) on \(\Omega\) such that \(f'(z) \neq 0\) everywhere.

## Directional Derivative

Source: Mathématiques III, Francis Maisonneuve, Presses des Mines.

### Questions

Let \(f\) be a complex-valued function defined in a neighbourhood of a point \(z_0 \in \mathbb{C}\). Assume that \(f\) is \(\mathbb{R}\)-differentiable at \(z_0\).Let \(\alpha \in \mathbb{R}\) and \(z_{r,\alpha} = z_0 + r e^{i\alpha}\) for \(r \in \mathbb{R}\). Show that \[ \ell_{\alpha} = \lim_{r \to 0} \frac{f(z_{r,\alpha}) - f(z_0)}{z_{r,\alpha} - z_0} \] exists and determine its value as a function of \(df_{z_0}\) and \(\alpha\).

What is the geometric structure of the set \(A = \{\ell_{\alpha} \; | \; \alpha \in \mathbb{R}\}\) ?

For which of these sets \(A\) is \(f\) \(\mathbb{C}\)-differentiable at \(z_0\)?

### Answers

The real-differentiability of \(f\) at \(z_0\) provides \[ f(z_0 + re^{i\alpha}) = f(z_0) + df_{z_0}(re^{i\alpha}) + \epsilon_{z_0}(re^{i\alpha}) |r| \] where \(\lim_{h \to 0} \epsilon_{z_0}(h) = 0\). Therefore, \[ \frac{f(z_{r,\alpha}) - f(z_0)}{z_{r,\alpha} - z_0} = (re^{i\alpha})^{-1}(df_{z_0}(re^{i\alpha}) + \epsilon_{z_0}(re^{i\alpha})|r|). \] Using the \(\mathbb{R}\)-linearity of \(df_{z_0}\), we get \[ \frac{f(z_{r,\alpha}) - f(z_0)}{z_{r,\alpha} - z_0} = e^{-i\alpha}df_{z_0}(e^{i\alpha}) + \epsilon_{z_0,\alpha}(r) \] for some function \(\epsilon_{z_0,\alpha}\) such that \(\lim_{r \to 0} \epsilon_{z_0,\alpha}(r) = 0\). Hence, the limit that defines \(\ell_{\alpha}\) exists and \[ \ell_{\alpha} = e^{-i\alpha}df_{z_0}(e^{i\alpha}). \]

For every real number \(\alpha\), we have \[ \ell_{\alpha} = e^{-i\alpha} \left( \frac{\partial f}{\partial x}(z_0) \cos \alpha + \frac{\partial f}{\partial y}(z_0) \sin \alpha \right). \] Hence, if we use the equations \[ \cos \alpha = \frac{e^{i\alpha} + e^{-i\alpha}}{2}, \; \sin \alpha = \frac{e^{i\alpha} - e^{-i\alpha}}{2i}, \; \] we obtain \[ \ell_{\alpha} = \frac{1}{2} \left( \frac{\partial f}{\partial x}(z_0) + \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \right) + \frac{1}{2} \left( \frac{\partial f}{\partial x}(z_0) - \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \right)e^{-i2\alpha}. \] Therefore, the set \(A = \{\ell_{\alpha} \; | \; \alpha \in \mathbb{R}\}\) is a circle centered on \[ c = \frac{\partial f}{\partial x}(z_0) + \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \] whose radius is \[ r = \left| \frac{\partial f}{\partial x}(z_0) - \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \right|. \]

The function \(f\) is complex-differentiable at \(z_0\) if and only if the Cauchy-Riemann equation \[ \frac{\partial f}{\partial x}(z_0) = \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \] is met, which happens exactly when the radius of the circle \(A\) is zero, that is, when \(A\) is a single point in the complex plane.