1. Every constant function \(f: z \in \mathbb{C} \mapsto \lambda \in \mathbb{C}\) is holomorphic as \[ \forall \, z \in \mathbb{C}, \, f'(z) = \lim_{h \to 0} \frac{\lambda - \lambda}{h} = 0. \]

2. The identity function \(f: z \in \mathbb{C} \mapsto z\) is holomorphic: \[ \forall \, z \in \mathbb{C}, \, f'(z) = \lim_{h \to 0} \frac{(z+h) - z}{h} = 1. \]

3. The inverse function \(f: z \in \mathbb{C}^* \to 1/z\) is holomorphic: the set \(\mathbb{C}^*\) is open and for any \(z \in \mathbb{C}^*\) and any \(h \in \mathbb{C}\) such that \(z+h \neq 0,\) we have \[ \frac{1/(z+h) - 1/z}{h} = -\frac{1}{z(z+h)}, \] hence \[ f'(z) = \lim_{h \to 0} -\frac{1}{z(z+h)} = - \frac{1}{z^2}. \]

4. The complex conjugate function \(f: z\in\mathbb{C} \to \overline{z}\) is nowhere complex-differentiable. Its difference quotient satisfies \[ \frac{f(z+h)-f(z)}{h} = \frac{\overline{z+h} - \overline{z}}{h} = \frac{\overline{h}}{h}, \] therefore when \(t \in \mathbb{R},\) \[ \lim_{t \to 0} \frac{f(z+t) - f(z)}{t} = +1, \] but \[ \lim_{t \to 0} \frac{f(z+it) - f(z)}{it} = -1, \] hence the difference quotient has no limit when \(h \to 0.\)

We mention this property now because we will use it to simplify the statements of some results of the current and subsequent chapters. Unfortunately, we cannot prove it yet; it is a consequence of Cauchy’s integral theory which is far from being trivial. To make sure that we won’t develop a circular argument, we flag the results that use this property with the symbol \([\dagger],\) until we can prove it.

A major result of complex analysis, Cauchy’s integral theorem, was originally formulated under the assumption that the derivative exists and is continuous² (Cauchy 1825). We have to wait for the paper “Sur la définition générale des fonctions analytiques, d’après Cauchy” (Goursat 1900) to officially get rid of this assumption:

J’ai reconnu depuis longtemps que la démonstration du théorème de Cauchy, que j’ai donnée en 1883, ne supposait pas la continuité de la dérivée. Pour répondre au désir qui m’a été exprimé par M. le Professeur W. F. Osgood, je vais indiquer ici rapidement comment on peut faire cette extension.

which means:

I have long recognized that the proof of Cauchy’s theorem, that I have given in 1883, did not assume the continuity of the derivative. To meet the desire which was expressed to me by Professor W. F. Osgood, I’ll tell here quickly how we can make this extension.

Because of this improvement, Cauchy’s integral theorem is also known as the “Cauchy-Goursat theorem”. Refer to (Hille 1973, footnote p.163) for a broader historical perspective on this subject.

Note how things change if we consider the plane as a real vector space: it is of dimension 2 with for example \(\{1,i\}\) as a basis. In particular, the vectors \(1\) and \(i\) which are complex-colinear are not real-colinear.

Proof.

hence the derivative of \(f \times g\) exists and satisfies the product rule. ‌\(\blacksquare\)

• it is real-differentiable on \(\Omega\) and

• for every \(z\) in \(\Omega,\) \(df_z\) is complex-linear.

The second clause may be replaced by any of the following:

1. the function \(f\) satisfies

   \[\forall\, z \in \Omega, \; df_z(i) = idf_z(1),\]

2. the function \(f\) satisfies the (complex) Cauchy-Riemann equation:

   \[ \frac{\partial f}{\partial x} = \frac{1}{i}\frac{\partial f}{\partial y}, \]

3. the functions \(u\) and \(v\) satisfy the (scalar) Cauchy-Riemann equations:

   \[ \frac{\partial u}{\partial x} = +\frac{\partial v}{\partial y} \; \mbox{ and } \; \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \]

If the complex-differentiability holds, we have \[ \begin{split} f' = (z \mapsto d_z f(1)) &= \frac{\partial f}{\partial x} = \frac{1}{i}\frac{\partial f}{\partial y} \\ &= \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y} \\ &= \frac{\partial u}{\partial x} - i \frac{\partial u}{\partial y} = \frac{\partial v}{\partial y} + i \frac{\partial v}{\partial x} \end{split} \]

Assume that \(f\) is real-differentiable; the real-differential \(\ell = df_z\) is real-linear, that is, additive and real-homogeneous. If additionally \(\ell(i) = i \ell(1),\) then we have for any real numbers \(\mu,\) \(\nu,\) \(x\) and \(y\)

• \(\ell(\mu x) = \mu\ell(x),\)

• \(\ell(i\nu x) = \nu x \ell(i) = \nu x i \ell(1) = i \nu \ell(x),\)

• \(\ell(i\mu y) = \mu\ell(iy),\)

• \(\ell(-\nu y) = - \nu y \ell(1) = i^2 \nu y \ell(1) = i \nu y \ell(i) = i \nu \ell(iy).\)

The function \(\ell\) is additive, hence \(\ell((\mu+i\nu)(x+iy)) = (\mu+i\nu) \ell(x+iy)\): the function \(\ell\) is complex-homogeneous and therefore complex-linear. Hence, property 1 yields the complex-linearity of the differential.

As additionally \[ \frac{\partial f}{\partial x}(z) = df_z(1), \; \frac{\partial f}{\partial y}(z) = df_z(i), \] properties 1 and 2 are equivalent.

The function \(f=(u,v)=u+iv\) is real-differentiable if and only if \(u\) and \(v\) are real-differentiable. In this case, \(df = (du, dv) = du + idv,\) hence \[ \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} , \; \frac{\partial f}{\partial y} = \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} \] which yields the equivalence between properties 2 and 3. ‌\(\blacksquare\)

1. The partial derivatives \(\partial f/\partial x\) and \(\partial f/\partial y\) exist, are continuous and

   \[ \frac{\partial f}{\partial x} = \frac{1}{i}\frac{\partial f}{\partial y}. \]

2. The partial derivatives \(\partial u / \partial x,\) \(\partial u / \partial y,\) \(\partial v / \partial x\) and \(\partial v / \partial y\) exist, are continuous and \[ \frac{\partial u}{\partial x} = +\frac{\partial v}{\partial y} \; \mbox{ and } \; \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \]

Reciprocally, if the derivative of \(f\) exist, then it is continuous, hence the partial derivatives of \(f\) (or of \(u\) and \(v\)) are continous. The previous theorem also shows that the Cauchy-Riemann equations are satisfied. ‌\(\blacksquare\)

\[ e^z = e^x \times (\cos y + i \sin y). \] The exponential function \(\exp: \mathbb{C} \to \mathbb{C}\) satisfies \[ \frac{\partial e^{x+iy}}{\partial x} = e^x \times (\cos y + i \sin y) = e^{x+iy} \] and \[ \frac{1}{i} \frac{\partial e^{x+iy}}{\partial y} = \frac{1}{i} e^x \times (-\sin y + i \cos y) = e^{x+iy}. \] Both partial derivatives exist and are continuous. Since they also satisfy the Cauchy-Riemann equations, the exponential function is complex-differentiable and \[ \frac{d e^z}{dz} = e^z. \]

It is a bijection from \(\mathbb{C} \setminus \mathbb{R}_-\) into \(\mathbb{R} \times \left]-\pi, +\pi \right[\) and for every \(z \in \mathbb{C} \setminus \mathbb{R}_-\) \[ e^{\log z} = \exp \circ \log (z) = z. \] The exponential function is continuously real-differentiable and \[d\exp_z (h) = (\exp z) \times h\] hence its differential \(d \exp_z\) is invertible. By the inverse function theorem, \(\log\) is real-differentiable on \(\mathbb{C} \setminus \mathbb{R}_-\) and \[ d \exp_{\log z} \circ \, d\log_z (h) = z \times d \log_z (h) = h. \] Consequently, \(d \log_z(h) = h/z,\) which is a complex-linear function of \(h.\) Hence, \(\log\) is complex-differentiable and \[ \log'(z) = \frac{1}{z}. \]