# Exercises

## Antiholomorphic Functions

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

### Questions

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

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

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

4. What property has the composition of two antiholomorphic functions?

5. 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

1. 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.

2. 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 $df_z = d(c\circ \overline{f})_z = c \circ d\overline{f}_z.$

3. 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 $\frac{\partial f}{\partial x}(z) = -\frac{1}{i}\frac{\partial f}{\partial y}(z), \; \mbox{ or } d f_z(i) = -i \times d f_z(1).$

4. 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.

5. 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

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

2. Show that the function $$\log$$ is holomorphic and compute its derivative.

### Answers

1. 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.

2. 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.
1. 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.

2. Identify the class of conformal mappings defined on $$\Omega$$.

### Answers

1. 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.

2. 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 $$z_0$$.
1. 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 $$\alpha$$.

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

3. For which of these sets $$A$$ is $$f$$ $$\mathbb{C}$$-differentiable at $$z_0$$?

### Answers

1. 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 $\ell_{\alpha} = e^{-i\alpha}df_{z_0}(e^{i\alpha}).$

2. 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 $r = \left| \frac{\partial f}{\partial x}(z_0) - \frac{1}{i}\frac{\partial f}{\partial y}(z_0) \right|.$

3. 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.