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complex.pdf page 15
Definition 3.10. Let $f: U \rightarrow \mathbb{C}$ be a function with components $(u, v)$, and suppose that the partial derivatives of these exist. Then we define the Wirtinger (partial) derivatives by$$\partial_{z} f:=\frac{1}{2}\left(\partial_{x}-i \partial_{y}\right) u+i \frac{1}{2}\left(\partial_{x}-i \partial_{y}\right) v$$and$$\partial_{\bar{z}} f:=\frac{1}{2}\left(\partial_{x}+i \partial_{y}\right) u+i \frac{1}{2}\left(\partial_{x}+i \partial_{y}\right) v$$Lemma 3.11. Let $U$ be an open subset of $\mathbb{C}$ and let $f: U \rightarrow \mathbb{C}$. Then $f$ satisfies the Cauchy-Riemann equations if and only if $\partial_{\bar{z}} f=0$.
Proof.\begin{aligned}\partial_{\bar{z}} f&=\frac{1}{2}\left(\partial_{x}+i \partial_{y}\right) u+i \frac{1}{2}\left(\partial_{x}+i \partial_{y}\right) v\\&=\frac{1}{2}\left(\partial_{x}u-\partial_{y}v\right) +i \frac{1}{2}\left(\partial_{y}u+\partial_{x}v\right) \\&=0+0i\end{aligned} |
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Basic properties$${\partial\over\partial z_i}z_i={\partial\over\partial \bar z_i}\bar z_i=1$$Linearity: For $f,g\in C^{1}(\Omega ),$\begin{aligned}{\frac {\partial }{\partial z_{i}}}\left(\alpha f+\beta g\right)&=\alpha {\frac {\partial f}{\partial z_{i}}}+\beta {\frac {\partial g}{\partial z_{i}}}\\{\frac {\partial }{\partial {\bar {z}}_{i}}}\left(\alpha f+\beta g\right)&=\alpha {\frac {\partial f}{\partial {\bar {z}}_{i}}}+\beta {\frac {\partial g}{\partial {\bar {z}}_{i}}}\end{aligned}Product rule: For $f,g\in C^{1}(\Omega ),$\begin{aligned}{\frac {\partial }{\partial z_{i}}}(f\cdot g)&={\frac {\partial f}{\partial z_{i}}}\cdot g+f\cdot {\frac {\partial g}{\partial z_{i}}}\\{\frac {\partial }{\partial {\bar {z}}_{i}}}(f\cdot g)&={\frac {\partial f}{\partial {\bar {z}}_{i}}}\cdot g+f\cdot {\frac {\partial g}{\partial {\bar {z}}_{i}}}\end{aligned}Chain rule: For $f,g\in C^{1}(\Omega ),g(\Omega)\subset\Omega,$\begin{aligned}{\frac {\partial }{\partial z}}(f\circ g)&=\left({\frac {\partial f}{\partial z}}\circ g\right){\frac {\partial g}{\partial z}}+\left({\frac {\partial f}{\partial {\bar {z}}}}\circ g\right){\frac {\partial {\bar {g}}}{\partial z}}\\{\frac {\partial }{\partial {\bar {z}}}}(f\circ g)&=\left({\frac {\partial f}{\partial z}}\circ g\right){\frac {\partial g}{\partial {\bar {z}}}}+\left({\frac {\partial f}{\partial {\bar {z}}}}\circ g\right){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}\end{aligned} |
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