§2.2 Complex Differentiation

2.2 Complex Differentiation

For $𝑈 \subseteq ℂ$ and a complex function $𝑓 : 𝑈 \to ℂ$ , $𝑓 (𝑧)$ is complex differentiable at $𝑧 \in 𝑈$ if the following limit exists, regardless of the direction $Δ 𝑧$ approaches $0$ from:

lim_{Δ 𝑧 \to 0} \frac{𝑓 (𝑧 + Δ 𝑧) - 𝑓 (𝑧)}{Δ 𝑧} .

We can consider $𝑓 (𝑧)$ to be a bivariate function $𝑓 (𝑥, 𝑦)$ for $𝑧 = 𝑥 + i 𝑦$ . Two main cases we are concerned with are when $Δ 𝑧$ approaches $0$ from the real and imaginary axes:

lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑧 + Δ 𝑧) - 𝑓 (𝑧)}{Δ 𝑧} = lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑧 + i Δ 𝑧) - 𝑓 (𝑧)}{i Δ 𝑧} .

Expressing $𝑓 (𝑧)$ as $𝑓 (𝑥, 𝑦) = 𝑢 (𝑥, 𝑦) + i 𝑣 (𝑥, 𝑦)$ ,

lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑧 + Δ 𝑧) - 𝑓 (𝑧)}{Δ 𝑧} = lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑥 + Δ 𝑧, 𝑦) - 𝑓 (𝑥, 𝑦)}{Δ 𝑧} = \frac{𝜕 𝑢}{𝜕 𝑥} + i \frac{𝜕 𝑣}{𝜕 𝑥},

and

lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑧 + i Δ 𝑧) - 𝑓 (𝑧)}{i Δ 𝑧} = - i lim_{\begin{matrix} Δ 𝑧 \to 0 \\ Δ 𝑧 \in ℝ \end{matrix}} \frac{𝑓 (𝑥, 𝑦 + Δ 𝑧) - 𝑓 (𝑥, 𝑦)}{Δ 𝑧} = \frac{𝜕 𝑣}{𝜕 𝑦} - i \frac{𝜕 𝑢}{𝜕 𝑦} .

By comparing the real and imaginary parts, we obtain necessary conditions for complex differentiability:

\frac{𝜕 𝑢}{𝜕 𝑥} = \frac{𝜕 𝑣}{𝜕 𝑦} and \frac{𝜕 𝑣}{𝜕 𝑥} = - \frac{𝜕 𝑢}{𝜕 𝑦} (2.2 . 1)

By multiplying the second equation by $i$ and adding it to the first, we obtain the equivalent form

\frac{𝜕 𝑓}{𝜕 𝑥} = - i \frac{𝜕 𝑓}{𝜕 𝑦} . (2.2 . 2)

The identities (2.2.1) $(2.2.1) (2.2.1)$ and (2.2.2) $(2.2.2) (2.2.2)$ are known as the Cauchy–Riemann equations.

Although this condition is necessary, it is not sufficient. Consider the function

𝑓 (𝑧) = \sqrt{| ℜ𝔢 (𝑧) ℑ𝔪 (𝑧) |} .

Let $𝑧 = 𝑥 + i 𝑦$ , $𝑥 = 𝛼 𝑡$ , and $𝑦 = 𝛽 𝑡$ . Then

lim_{𝑧 \to 0} \frac{𝑓 (𝑧) - 𝑓 (0)}{𝑧 - 0} = lim_{𝑧 \to 0} \frac{𝑓 (𝑧)}{𝑧} = lim_{𝑡 \to 0} \frac{\sqrt{| 𝛼 𝛽 𝑡^{2} |}}{𝛼 𝑡 + i 𝛽 𝑡} = \frac{\sqrt{| 𝛼 𝛽 |}}{𝛼 + i 𝛽} .

The derivative along $𝛼 = 1$ , $𝛽 = 0$ , or the real axis, vanishes. Along $𝛼 = 0$ , $𝛽 = 1$ , or the imaginary axis, it also vanishes. However, the limit is different for any other pair of $𝛼$ and $𝛽$ , and hence for other directions of approach.

Definition 2.2.1 (Holomorphy).

A function

𝑓 : 𝑈 \to ℂ

is said to be holomorphic at

𝑧_{0} \in 𝑈

if it is complex differentiable on a neighborhood of

𝑧_{0}

. If

𝑓 (𝑧)

is holomorphic for every point in an open connected set

𝑈

, then it is said to be holomorphic over

𝑈

. A function is holomorphic over a compact set

𝐾

if it is holomorphic on a neighborhood of

𝐾

Weierstrass provided the following classification:

Definition 2.2.2.

A function is entire if it is holomorphic over

ℂ

For the purpose of the following contents, a region or domain will denote a nonempty, open, connected subset of the complex plane.

Theorem 2.2.3.

Let

𝑈 \subseteq ℂ

be open, and let

𝑓 : 𝑈 \to ℂ

be a function. Then

𝑓

is holomorphic on

𝑈

iff

𝑓 \in 𝐶^{1} (𝑈)

and satisfies the Cauchy–Riemann equations.

Proof.

The first part is to prove that any holomorphic function on $𝑈$ has continuous first-order partial derivatives in $𝑈$ . This requires an argument that will be covered later, specifically in Section 3.2 $Section 3.2 Section 3.2$ , which states that the complex derivative of any holomorphic function is also holomorphic over the region.

For the second part, let $𝑓 (𝑧) = 𝑓 (𝑥, 𝑦) = 𝑢 (𝑥, 𝑦) + i 𝑣 (𝑥, 𝑦)$ . Assume that $𝑢, 𝑣 \in 𝐶^{1} ({𝑧_{0}})$ and satisfy the Cauchy–Riemann equations at $𝑧_{0} = 𝑥_{0} + i 𝑦_{0}$ . Let

𝛼 = \frac{𝜕 𝑢}{𝜕 𝑥} (𝑥_{0}, 𝑦_{0}) = \frac{𝜕 𝑣}{𝜕 𝑦} (𝑥_{0}, 𝑦_{0}), 𝛽 = \frac{𝜕 𝑣}{𝜕 𝑥} (𝑥_{0}, 𝑦_{0}) = - \frac{𝜕 𝑢}{𝜕 𝑦} (𝑥_{0}, 𝑦_{0}) .

Then because $𝑢, 𝑣 \in 𝐶^{1} (𝑈)$ have continuous partial derivatives, $\forall 𝑥 + i 𝑦 \in 𝑈$ :

𝑢 (𝑥, 𝑦) - 𝑢 (𝑥_{0}, 𝑦_{0}) = 𝛼 (𝑥 - 𝑥_{0}) - 𝛽 (𝑦 - 𝑦_{0}) + ℴ︀ (| Δ 𝑧 |),

𝑣 (𝑥, 𝑦) - 𝑣 (𝑥_{0}, 𝑦_{0}) = 𝛽 (𝑥 - 𝑥_{0}) + 𝛼 (𝑦 - 𝑦_{0}) + ℴ︀ (| Δ 𝑧 |),

where $| Δ 𝑧 | = \sqrt{Δ 𝑥^{2} + Δ 𝑦^{2}}$ and $ℴ︀ (| Δ 𝑧 |)$ denotes a value with higher infinitesimal order to $| Δ 𝑧 |$ , or that ${lim}_{Δ 𝑧 \to 0} \frac{ℴ︀ (| Δ 𝑧 |)}{| Δ 𝑧 |} = 0$ . Then letting $Δ 𝑧 = 𝑥 - 𝑥_{0} + i (𝑦 - 𝑦_{0})$ ,

𝑓 (𝑧) - 𝑓 (𝑧_{0}) = 𝛼 Δ 𝑧 + i 𝛽 Δ 𝑧 + ℴ︀ (| Δ 𝑧 |) + ℴ︀ (| Δ 𝑧 |),

\frac{𝑓 (𝑧) - 𝑓 (𝑧_{0})}{𝑧 - 𝑧_{0}} = 𝛼 + i 𝛽 + \frac{ℴ︀ (| Δ 𝑧 |)}{| Δ 𝑧 |} \cdot \frac{| Δ 𝑧 |}{Δ 𝑧} .

Taking the limit as $Δ 𝑧 \to 0$ , the high order infinitesimals on the right-hand side vanish, and

lim_{Δ 𝑧 \to 0} \frac{𝑓 (𝑧) - 𝑓 (𝑧_{0})}{𝑧 - 𝑧_{0}} = 𝛼 + i 𝛽 . □

We will prove later in Section 3.2 $Section 3.2 Section 3.2$ that the complex derivative of a holomorphic function $𝑓 (𝑧) = 𝑢 (𝑧) + i 𝑣 (𝑧)$ is holomorphic. Under this assumption, $𝑓 (𝑧)$ has continuous second-order partial derivatives, and therefore,

\frac{𝜕^{2} 𝑢}{𝜕 𝑥 𝜕 𝑦} = \frac{𝜕^{2} 𝑢}{𝜕 𝑦 𝜕 𝑥}, \frac{𝜕^{2} 𝑣}{𝜕 𝑥 𝜕 𝑦} = \frac{𝜕^{2} 𝑣}{𝜕 𝑦 𝜕 𝑥},

and by the Cauchy–Riemann equations,

\frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} = \frac{𝜕^{2} 𝑣}{𝜕 𝑥 𝜕 𝑦}, \frac{𝜕^{2} 𝑢}{𝜕 𝑦^{2}} = - \frac{𝜕^{2} 𝑣}{𝜕 𝑦 𝜕 𝑥},

and

\frac{𝜕^{2} 𝑣}{𝜕 𝑥^{2}} = - \frac{𝜕^{2} 𝑢}{𝜕 𝑥 𝜕 𝑦}, \frac{𝜕^{2} 𝑣}{𝜕 𝑦^{2}} = \frac{𝜕^{2} 𝑢}{𝜕 𝑦 𝜕 𝑥} .

Adding the equations,

\frac{𝜕^{2} 𝑢}{𝜕 𝑥^{2}} + \frac{𝜕^{2} 𝑢}{𝜕 𝑦^{2}} = 0, \frac{𝜕^{2} 𝑣}{𝜕 𝑥^{2}} + \frac{𝜕^{2} 𝑣}{𝜕 𝑦^{2}} = 0 .

This general type of equation is known as Laplace’s equation, which is a basic example of an elliptic partial differential equation. Define the operator (the Laplacian)

\nabla^{2} = 𝛁 \cdot 𝛁 = \frac{𝜕^{2}}{𝜕 𝑥^{2}} + \frac{𝜕^{2}}{𝜕 𝑦^{2}} .

Definition 2.2.4 (Harmonicity).

For a given region

𝑈 \subseteq ℂ

, a real-valued function

𝑢 \in 𝐶^{2} (𝑈)

satisfying Laplace’s equation

\nabla^{2} 𝑢 = 0

is a harmonic function.

Thus, the real and complex parts of a holomorphic function are harmonic functions.

Letting $𝑥 = 𝑟 cos 𝜃$ , $𝑦 = 𝑟 sin 𝜃$ , the Laplacian is equal to:

\begin{aligned} \nabla^{2} & = \frac{𝜕^{2}}{𝜕 𝑥^{2}} + \frac{𝜕^{2}}{𝜕 𝑦^{2}} = \frac{𝜕}{𝜕 𝑥} (\frac{𝜕 𝑟}{𝜕 𝑥} \frac{𝜕}{𝜕 𝑟} + \frac{𝜕 𝜃}{𝜕 𝑥} \frac{𝜕}{𝜕 𝜃}) + \frac{𝜕}{𝜕 𝑦} (\frac{𝜕 𝑟}{𝜕 𝑦} \frac{𝜕}{𝜕 𝑟} + \frac{𝜕 𝜃}{𝜕 𝑦} \frac{𝜕}{𝜕 𝜃}) \\ = \frac{𝜕}{𝜕 𝑥} (\frac{𝑥}{𝑟} \frac{𝜕}{𝜕 𝑟} - \frac{𝑦}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) + \frac{𝜕}{𝜕 𝑦} (\frac{𝑦}{𝑟} \frac{𝜕}{𝜕 𝑟} + \frac{𝑥}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) \\ = (\frac{𝑥}{𝑟} \frac{𝜕}{𝜕 𝑟} - \frac{𝑦}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) (\frac{𝑥}{𝑟} \frac{𝜕}{𝜕 𝑟} - \frac{𝑦}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) + (\frac{𝑦}{𝑟} \frac{𝜕}{𝜕 𝑟} + \frac{𝑥}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) (\frac{𝑦}{𝑟} \frac{𝜕}{𝜕 𝑟} + \frac{𝑥}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃}) \\ = (cos 𝜃 \frac{𝜕}{𝜕 𝑟} - \frac{sin 𝜃}{𝑟} \frac{𝜕}{𝜕 𝜃}) (cos 𝜃 \frac{𝜕}{𝜕 𝑟} - \frac{sin 𝜃}{𝑟} \frac{𝜕}{𝜕 𝜃}) \\ \begin{matrix}  \end{matrix} + (sin 𝜃 \frac{𝜕}{𝜕 𝑟} + \frac{cos 𝜃}{𝑟} \frac{𝜕}{𝜕 𝜃}) (sin 𝜃 \frac{𝜕}{𝜕 𝑟} + \frac{cos 𝜃}{𝑟} \frac{𝜕}{𝜕 𝜃}) \\ = {cos}^{2} 𝜃 \frac{𝜕^{2}}{𝜕 𝑟^{2}} - cos 𝜃 sin 𝜃 (- \frac{1}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃} + \frac{1}{𝑟} \frac{𝜕^{2}}{𝜕 𝜃 𝜕 𝑟}) \\ \begin{matrix}  \end{matrix} - \frac{sin 𝜃}{𝑟} (- sin 𝜃 \frac{𝜕}{𝜕 𝑟} + cos 𝜃 \frac{𝜕^{2}}{𝜕 𝜃 𝜕 𝑟}) + \frac{sin 𝜃}{𝑟^{2}} (cos 𝜃 \frac{𝜕}{𝜕 𝜃} + sin 𝜃 \frac{𝜕^{2}}{𝜕 𝜃^{2}}) \\ \begin{matrix}  \end{matrix} + {sin}^{2} 𝜃 \frac{𝜕^{2}}{𝜕 𝑟^{2}} + sin 𝜃 cos 𝜃 (- \frac{1}{𝑟^{2}} \frac{𝜕}{𝜕 𝜃} + \frac{1}{𝑟} \frac{𝜕^{2}}{𝜕 𝜃 𝜕 𝑟}) \\ \begin{matrix}  \end{matrix} + \frac{cos 𝜃}{𝑟} (cos 𝜃 \frac{𝜕}{𝜕 𝑟} + sin 𝜃 \frac{𝜕^{2}}{𝜕 𝜃 𝜕 𝑟}) + \frac{cos 𝜃}{𝑟^{2}} (- sin 𝜃 \frac{𝜕}{𝜕 𝜃} + cos 𝜃 \frac{𝜕^{2}}{𝜕 𝜃^{2}}) \\ = \frac{𝜕^{2}}{𝜕 𝑟^{2}} + \frac{1}{𝑟^{2}} \frac{𝜕^{2}}{𝜕 𝜃^{2}} + \frac{1}{𝑟} \frac{𝜕}{𝜕 𝑟} . (2.2 . 3) \end{aligned}

Proposition 2.2.5.

Let

𝑈 \subseteq ℂ

be open and connected and let

𝑓 : 𝑈 \to ℝ

be holomorphic. It follows that

𝑓

is constant over

𝑈

Proof.

Since

𝑓 (𝑥, 𝑦) = 𝑢 (𝑥, 𝑦) + i 𝑣 (𝑥, 𝑦)

is real-valued,

𝑣 (𝑥, 𝑦) \equiv 0

𝑈

. Then by the Cauchy–Riemann equations on

𝑈

\frac{𝜕 𝑢}{𝜕 𝑥} = \frac{𝜕 𝑣}{𝜕 𝑦} = 0

. Similarly,

\frac{𝜕 𝑢}{𝜕 𝑦} = - \frac{𝜕 𝑣}{𝜕 𝑥} = 0 .

Therefore,

𝑓 (𝑧) = 𝑢 (𝑧)

is constant. ⁠

□