§2.2.1 Wirtinger Derivatives

2.2.1 Wirtinger Derivatives

We have previously introduced the concept of expressing a complex function as a function of $𝑥$ and $𝑦$ . It can also be expressed in terms of $𝑧$ and $\overline{𝑧}$ , where $𝑧 = 𝑥 + i 𝑦$ and $\overline{𝑧} = 𝑥 - i 𝑦$ . Then ${| 𝑧 |}^{2} = 𝑧 \overline{𝑧}$ , $𝑥 = \frac{𝑧 + \overline{𝑧}}{2}$ , and $𝑦 = \frac{𝑧 - \overline{𝑧}}{2 i}$ . By the rules of the derivative, it is only natural that we define

\frac{𝜕}{𝜕 𝑧} = \frac{𝜕}{𝜕 𝑥} \frac{𝜕 𝑥}{𝜕 𝑧} + \frac{𝜕}{𝜕 𝑦} \frac{𝜕 𝑦}{𝜕 𝑧} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} - i \frac{𝜕}{𝜕 𝑦}) (2.2.1 . 1)

and

\frac{𝜕}{𝜕 \overline{𝑧}} = \frac{𝜕}{𝜕 𝑥} \frac{𝜕 𝑥}{𝜕 \overline{𝑧}} + \frac{𝜕}{𝜕 𝑦} \frac{𝜕 𝑦}{𝜕 \overline{𝑧}} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} + i \frac{𝜕}{𝜕 𝑦}) . (2.2.1 . 2)

If (2.2.1.1) $(2.2.1.4) (2.2.1.1)$ is set equal to $0$ , then it is the equivalent form of the homogeneous Cauchy–Riemann Equations. Then for a holomorphic function $𝑓 (𝑧)$ , the Wirtinger derivative $\frac{𝜕 𝑓}{𝜕 𝑧} = \frac{d 𝑓}{d 𝑧}$ .

In terms of $𝑢$ and $𝑣$ , the two derivatives of a function $𝑓 (𝑧)$ are equal to:

\frac{𝜕 𝑓}{𝜕 𝑧} = \frac{1}{2} (\frac{𝜕 𝑢}{𝜕 𝑥} + i \frac{𝜕 𝑣}{𝜕 𝑥} - i \frac{𝜕 𝑢}{𝜕 𝑦} + \frac{𝜕 𝑣}{𝜕 𝑦}),

and

\frac{𝜕 𝑓}{𝜕 \overline{𝑧}} = \frac{1}{2} (\frac{𝜕 𝑢}{𝜕 𝑥} + i \frac{𝜕 𝑣}{𝜕 𝑥} + i \frac{𝜕 𝑢}{𝜕 𝑦} - \frac{𝜕 𝑣}{𝜕 𝑦}) .

If $𝑓$ is holomorphic,

\frac{d 𝑓}{d 𝑧} = \frac{𝜕 𝑢}{𝜕 𝑥} + i \frac{𝜕 𝑣}{𝜕 𝑥} = \frac{𝜕 𝑣}{𝜕 𝑦} + i \frac{𝜕 𝑣}{𝜕 𝑥} = \frac{𝜕 𝑢}{𝜕 𝑥} - i \frac{𝜕 𝑢}{𝜕 𝑦} = \frac{𝜕 𝑣}{𝜕 𝑦} - i \frac{𝜕 𝑢}{𝜕 𝑦} . (2.2.1 . 3)

On the contrary, by the rules of the derivative,

\frac{𝜕}{𝜕 𝑥} = \frac{𝜕}{𝜕 𝑧} \frac{𝜕 𝑧}{𝜕 𝑥} + \frac{𝜕}{𝜕 \overline{𝑧}} \frac{𝜕 \overline{𝑧}}{𝜕 𝑥} = \frac{𝜕}{𝜕 𝑧} + \frac{𝜕}{𝜕 \overline{𝑧}}

and

\frac{𝜕}{𝜕 𝑦} = \frac{𝜕}{𝜕 𝑧} \frac{𝜕 𝑧}{𝜕 𝑦} + \frac{𝜕}{𝜕 \overline{𝑧}} \frac{𝜕 \overline{𝑧}}{𝜕 𝑦} = i \frac{𝜕}{𝜕 𝑧} - i \frac{𝜕}{𝜕 \overline{𝑧}} .

The Laplacian is equal to

\begin{aligned} Δ = \frac{𝜕^{2}}{𝜕 𝑥^{2}} + \frac{𝜕^{2}}{𝜕 𝑦^{2}} & = {(\frac{𝜕}{𝜕 𝑧} + \frac{𝜕}{𝜕 \overline{𝑧}})}^{2} + {(i \frac{𝜕}{𝜕 𝑧} - i \frac{𝜕}{𝜕 \overline{𝑧}})}^{2} \\ = \frac{𝜕^{2}}{𝜕 𝑧^{2}} + \frac{𝜕^{2}}{𝜕 {\overline{𝑧}}^{2}} + 2 \frac{𝜕^{2}}{𝜕 𝑧 𝜕 \overline{𝑧}} \\ \begin{matrix}  \end{matrix} - \frac{𝜕^{2}}{𝜕 𝑧^{2}} - \frac{𝜕^{2}}{𝜕 {\overline{𝑧}}^{2}} + 2 \frac{𝜕^{2}}{𝜕 𝑧 𝜕 \overline{𝑧}} \\ = 4 \frac{𝜕^{2}}{𝜕 𝑧 𝜕 \overline{𝑧}} . (2.2.1 . 4) \end{aligned}

Under this definition, we can derive the chain rule:

Theorem 2.2.1.1 (Chain Rule).

Let $Ω \subseteq ℂ$ be a region such that $𝑔 \in 𝐶^{1} (Ω)$ and $𝑓 \in 𝐶^{1} (𝑔 (Ω))$ . Writing $𝑤 = 𝑔 (𝑧)$ , it follows that

\begin{aligned} \frac{𝜕}{𝜕 𝑧} (𝑓 \circ 𝑔) & = (\frac{𝜕 𝑓}{𝜕 𝑤} \circ 𝑔) \frac{𝜕 𝑔}{𝜕 𝑧} + (\frac{𝜕 𝑓}{𝜕 \overline{𝑤}} \circ 𝑔) \frac{𝜕 \overline{𝑔}}{𝜕 𝑧} \\ \frac{𝜕}{𝜕 \overline{𝑧}} (𝑓 \circ 𝑔) & = (\frac{𝜕 𝑓}{𝜕 𝑤} \circ 𝑔) \frac{𝜕 𝑔}{𝜕 \overline{𝑧}} + (\frac{𝜕 𝑓}{𝜕 \overline{𝑤}} \circ 𝑔) \frac{𝜕 \overline{𝑔}}{𝜕 \overline{𝑧}} . \end{aligned}

Proof.

Write $𝑧 = 𝑥 + i 𝑦$ . Let

𝑔 (𝑧) = 𝜉 (𝑥, 𝑦) + i 𝜂 (𝑥, 𝑦), 𝑤 = 𝜉 + i 𝜂

so that $𝑤 = 𝑔 (𝑧)$ with $𝜉 = 𝜉 (𝑥, 𝑦), 𝜂 = 𝜂 (𝑥, 𝑦)$ . Let $𝑓$ be regarded as a $𝐶^{1}$ function of the real variables $𝜉, 𝜂$ ; equivalently we may view $𝑓$ as $𝑓 (𝑤, \overline{𝑤})$ where $\overline{𝑤} = 𝜉 - i 𝜂$ . The composition is $ℎ (𝑧) = 𝑓 \circ 𝑔 (𝑧) = 𝑓 (𝜉 (𝑥, 𝑦), 𝜂 (𝑥, 𝑦))$ .

Using the real chain rule (provided by the continuous differentiability), we have

\frac{𝜕 ℎ}{𝜕 𝑥} = \frac{𝜕 𝑓}{𝜕 𝜉} \frac{𝜕 𝜉}{𝜕 𝑥} + \frac{𝜕 𝑓}{𝜕 𝜂} \frac{𝜕 𝜂}{𝜕 𝑥}, \frac{𝜕 ℎ}{𝜕 𝑦} = \frac{𝜕 𝑓}{𝜕 𝜉} \frac{𝜕 𝜉}{𝜕 𝑦} + \frac{𝜕 𝑓}{𝜕 𝜂} \frac{𝜕 𝜂}{𝜕 𝑦} .

Hence,

\frac{𝜕 ℎ}{𝜕 𝑧} = \frac{1}{2} [\frac{𝜕 𝑓}{𝜕 𝜉} (\frac{𝜕 𝜉}{𝜕 𝑥} - i \frac{𝜕 𝜉}{𝜕 𝑦}) + \frac{𝜕 𝑓}{𝜕 𝜂} (\frac{𝜕 𝜂}{𝜕 𝑥} - i \frac{𝜕 𝜂}{𝜕 𝑦})] .

Now recall

\frac{𝜕 𝑓}{𝜕 𝑤} = \frac{1}{2} (\frac{𝜕 𝑓}{𝜕 𝜉} - i \frac{𝜕 𝑓}{𝜕 𝜂}), \frac{𝜕 𝑓}{𝜕 \overline{𝑤}} = \frac{1}{2} (\frac{𝜕 𝑓}{𝜕 𝜉} + i \frac{𝜕 𝑓}{𝜕 𝜂}) .

Thus,

\frac{𝜕 𝑓}{𝜕 𝜉} = \frac{𝜕 𝑓}{𝜕 𝑤} + \frac{𝜕 𝑓}{𝜕 \overline{𝑤}}, \frac{𝜕 𝑓}{𝜕 𝜂} = i (\frac{𝜕 𝑓}{𝜕 𝑤} - \frac{𝜕 𝑓}{𝜕 \overline{𝑤}}) .

Then by substitution,

\begin{aligned} \frac{𝜕 ℎ}{𝜕 𝑧} & = \frac{1}{2} [(\frac{𝜕 𝑓}{𝜕 𝑤} + \frac{𝜕 𝑓}{𝜕 \overline{𝑤}}) (\frac{𝜕 𝜉}{𝜕 𝑥} - i \frac{𝜕 𝜉}{𝜕 𝑦}) \\ \begin{matrix}  \end{matrix} + i (\frac{𝜕 𝑓}{𝜕 𝑤} - \frac{𝜕 𝑓}{𝜕 \overline{𝑤}}) (\frac{𝜕 𝜂}{𝜕 𝑥} - i \frac{𝜕 𝜂}{𝜕 𝑦})] \\ = \frac{𝜕 𝑓}{𝜕 𝑤} \frac{1}{2} [(\frac{𝜕 𝜉}{𝜕 𝑥} - i \frac{𝜕 𝜉}{𝜕 𝑦}) + i (\frac{𝜕 𝜂}{𝜕 𝑥} - i \frac{𝜕 𝜂}{𝜕 𝑦})] \\ \begin{matrix}  \end{matrix} + \frac{𝜕 𝑓}{𝜕 \overline{𝑤}} \frac{1}{2} [(\frac{𝜕 𝜉}{𝜕 𝑥} - i \frac{𝜕 𝜉}{𝜕 𝑦}) - i (\frac{𝜕 𝜂}{𝜕 𝑥} - i \frac{𝜕 𝜂}{𝜕 𝑦})] . \end{aligned}

The terms in brackets equal $\frac{𝜕 𝑔}{𝜕 𝑧}$ and $\frac{𝜕 \overline{𝑔}}{𝜕 𝑧}$ . Thus,

\frac{𝜕 ℎ}{𝜕 𝑧} = (\frac{𝜕 𝑓}{𝜕 𝑤} \circ 𝑔) \frac{𝜕 𝑔}{𝜕 𝑧} + (\frac{𝜕 𝑓}{𝜕 \overline{𝑤}} \circ 𝑔) \frac{𝜕 \overline{𝑔}}{𝜕 𝑧} .

A similar calculation using (2.2.1.2) $(2.2.1.5) (2.2.1.2)$ gives

\frac{𝜕}{𝜕 \overline{𝑧}} (𝑓 \circ 𝑔) = (\frac{𝜕 𝑓}{𝜕 𝑤} \circ 𝑔) \frac{𝜕 𝑔}{𝜕 \overline{𝑧}} + (\frac{𝜕 𝑓}{𝜕 \overline{𝑤}} \circ 𝑔) \frac{𝜕 \overline{𝑔}}{𝜕 \overline{𝑧}} .

These are exactly the proclaimed identities. ⁠ $□$

Last we have taking derivatives of conjugates:

Theorem 2.2.1.2.

Let $𝑓 \in 𝐶^{1} (Ω)$ where $Ω \subseteq ℂ$ is a region. Then

\frac{𝜕 \overline{𝑓}}{𝜕 𝑧} = \overline{(\frac{𝜕 𝑓}{𝜕 \overline{𝑧}})}, \frac{𝜕 \overline{𝑓}}{𝜕 \overline{𝑧}} = \overline{(\frac{𝜕 𝑓}{𝜕 𝑧})} .

Proof.

Write $𝑧 = 𝑥 + i 𝑦$ and $𝑓 (𝑧) = 𝑢 (𝑥, 𝑦) + i 𝑣 (𝑥, 𝑦)$ with $𝑢, 𝑣 \in 𝐶^{1} (Ω)$ . Then $\overline{𝑓 (𝑧)} = 𝑢 (𝑥, 𝑦) - i 𝑣 (𝑥, 𝑦)$ . We compute

\frac{𝜕 \overline{𝑓}}{𝜕 𝑧} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} - i \frac{𝜕}{𝜕 𝑦}) (𝑢 - i 𝑣) = \frac{1}{2} (𝑢_{𝑥}^{'} - 𝑣_{𝑦}^{'} - i (𝑣_{𝑥}^{'} + 𝑢_{𝑦}^{'})) .

On the other hand,

\frac{𝜕 𝑓}{𝜕 \overline{𝑧}} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} + i \frac{𝜕}{𝜕 𝑦}) (𝑢 + i 𝑣) = \frac{1}{2} (𝑢_{𝑥}^{'} - 𝑣_{𝑦}^{'} + i (𝑣_{𝑥}^{'} + 𝑢_{𝑦}^{'})) .

Taking complex conjugates yields

\overline{\frac{𝜕 𝑓}{𝜕 \overline{𝑧}}} = \frac{1}{2} (𝑢_{𝑥}^{'} - 𝑣_{𝑦}^{'} - i (𝑣_{𝑥}^{'} + 𝑢_{𝑦}^{'})) = \frac{𝜕 \overline{𝑓}}{𝜕 𝑧} .

Similarly,

\frac{𝜕 \overline{𝑓}}{𝜕 \overline{𝑧}} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} + i \frac{𝜕}{𝜕 𝑦}) (𝑢 - i 𝑣) = \frac{1}{2} (𝑢_{𝑥}^{'} + 𝑣_{𝑦}^{'} + i (𝑢_{𝑦}^{'} - 𝑣_{𝑥}^{'})),

while

\frac{𝜕 𝑓}{𝜕 𝑧} = \frac{1}{2} (\frac{𝜕}{𝜕 𝑥} - i \frac{𝜕}{𝜕 𝑦}) (𝑢 + i 𝑣) = \frac{1}{2} (𝑢_{𝑥}^{'} + 𝑣_{𝑦}^{'} + i (𝑣_{𝑥}^{'} - 𝑢_{𝑦}^{'})) .

Taking complex conjugates gives

\overline{\frac{𝜕 𝑓}{𝜕 𝑧}} = \frac{1}{2} (𝑢_{𝑥}^{'} + 𝑣_{𝑦}^{'} + i (𝑢_{𝑦}^{'} - 𝑣_{𝑥}^{'})) = \frac{𝜕 \overline{𝑓}}{𝜕 \overline{𝑧}} . □

As a final remark, we establish that the meaning of $𝐶^{𝑘} (Ω)$ does not refer to the existence of a $𝑘$ -th order complex derivative, but rather the existence of continuous partial derivatives up to order $𝑘$ in the real variables $𝑥$ and $𝑦$ (or equivalently, in terms of $𝑧$ and $\overline{𝑧}$ ).