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Let $\Omega \subset \mathbb{C}$ be an open set and $f=u+i v: \Omega \rightarrow \mathbb{C}$ be a real-differentiable function. Thinking of $\Omega$ as a subset of $\mathbb{R}^{2}$ and $f$ as a map $f=(u, v): \Omega \rightarrow \mathbb{R}^{2}$, the real derivative of $f$ is given by its Jacobian matrix $J_{f}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$,
$$
J_{f}=\left[\begin{array}{ll}
u_{x} & u_{y} \\
v_{x} & v_{y}
\end{array}\right] .
$$
If $f$ is holomorphic then the partial derivatives of $u$ and $v$ satisfy the Cauchy-Riemann equations $u_{x}=v_{y}, u_{y}=-v_{x}$, in which case we have
$$
J_{f}=\left[\begin{array}{cc}
u_{x} & u_{y} \\
-u_{y} & u_{x}
\end{array}\right] .
$$
In either case, $J_{f}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ is a real-linear map, meaning that $J_{f}\left(\lambda_{1} H_{1}+\lambda_{2} H_{2}\right)=$ $\lambda_{1} J_{f}\left(H_{1}\right)+\lambda_{2} J_{f}\left(H_{2}\right)$ for all $H_{1}, H_{2} \in \mathbb{R}^{2}, \lambda_{1}, \lambda_{2} \in \mathbb{R}$. But since $\mathbb{C}$ is just $\mathbb{R}^{2}$ with the additional structure of multiplication, we can ask when the derivative of $f$, thought of as a map $J_{f}: \mathbb{C} \rightarrow \mathbb{C}$, is complex-linear. That is, when do we have
$$
J_{f}\left(\lambda_{1} h_{1}+\lambda_{2} h_{2}\right)=\lambda_{1} J_{f}\left(h_{1}\right)+\lambda_{2} J_{f}\left(h_{2}\right) \text { for all } h_{1}, h_{2} \in \mathbb{C}, \lambda_{1}, \lambda_{2} \in \mathbb{C} \text { ? }
$$
Since $J_{f}$ is already real-linear, this reduces to asking when $J_{f}(i h)=i J_{f}(h)$ for all $h \in \mathbb{C}$.
If $h=h_{1}+i h_{2}=\left(h_{1}, h_{2}\right)$, then $i h=-h_{2}+i h_{1}=\left(-h_{2}, h_{1}\right)$. Then if $f$ is holomorphic we have
$$J_{f}(i h)=J_{f}\left(-h_{2}, h_{1}\right)=\left[\begin{array}{cc}u_{x} & u_{y} \\ -u_{y} & u_{x}\end{array}\right]\left[\begin{array}{c}-h_{2} \\ h_{1}\end{array}\right]=\left[\begin{array}{c}-u_{x} h_{2}+u_{y} h_{1} \\ u_{y} h_{2}+u_{x} h_{1}\end{array}\right]=i\left[\begin{array}{c}u_{x} h_{1}+u_{y} h_{2} \\ -u_{y} h_{1}+u_{x} h_{2}\end{array}\right]=i J_{f}(h)$$
showing that $J_{f}$ is complex-linear. Conversely, if $J_{f}$ is complex-linear then in particular for $h=1=(1,0)$ we have $J_{f}(i(1,0))=i J_{f}(1,0)$. But
$$
J_{f}(i(1,0))=J_{f}(0,1)=\left[\begin{array}{l}
u_{y} \\
v_{y}
\end{array}\right]
$$
and
$$
i J_{f}(1,0)=i\left[\begin{array}{l}
u_{x} \\
v_{x}
\end{array}\right]=\left[\begin{array}{c}
-v_{x} \\
u_{x}
\end{array}\right]
$$
which gives the Cauchy-Riemann equations
$$
\left[\begin{array}{l}
u_{y} \\
v_{y}
\end{array}\right]=\left[\begin{array}{c}
-v_{x} \\
u_{x}
\end{array}\right]
$$
showing that $f$ is holomorphic.
Thus, $f$ is holomorphic if and only if $J_{f}$ is complex-linear.
So, for $f: \Omega \rightarrow \mathbb{C}, f=u+i v$, the following are equivalent.
- $f$ is holomorphic (complex differentiable), that is, $f^{\prime}(z)$ exists for all $z \in \Omega$.
- $f$ is real-differentiable and $u, v$ satisfy the Cauchy-Riemann equations on $\Omega$.
- $f$ is real-differentiable and $J_{f}$ is complex linear at every point of $\Omega$.
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