Differentiation of Functions of a Complex Variable 1: Cauchy-Riemann Conditions

 Consider a complex-valued function $f(z)$ of a complex variable. The derivative $f'(z)$ is defined as
\begin{equation}
\label{eq:diff}
f'(z)=\lim_{\Delta z\to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}
\end{equation}
$f'(z)$ is also denoted by $\frac{df}{dz}$ . The limit in the above definition is independent of the particular approach to the point $z$. Suppose that $f'(z)$ exists. Let $z=x+iy$ and $f(z)=u(z)+iv(z)$. Then
$$\frac{\Delta f}{\Delta z}=\frac{\Delta u+i\Delta v}{\Delta x+i\Delta y}$$
First, let $\Delta y=0$ and $\Delta x\to 0$. Then
\begin{align*}
\lim_{\Delta z\to 0}\frac{\Delta f}{\Delta z}&=\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta z}+i\frac{\Delta v}{\Delta x}\\
&=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}
\end{align*}
This time, we let $\Delta x=0$ and $\Delta y\to 0$. Then
\begin{align*}
\lim_{\Delta z\to 0}\frac{\Delta f}{\Delta z}&=\lim_{\Delta y\to 0}\frac{\Delta u+i\Delta v}{i\Delta y}\\
&=\lim_{\Delta y\to 0}\frac{\Delta v}{\Delta y}-i\frac{\Delta u}{\Delta y}\\
&=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}
\end{align*}
Since $f'(z)$ exists, the two limits must coincide.
$$\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$$
Hence, we obtain
\begin{equation}
\label{eq:cr}
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}
\end{equation}
This is called the Cauchy-Riemann conditions. Conversely, let us assume that Cauchy-Riemann conditions are satisfied. In addition, we also assume that the partial derivatives of $u(x,y)$ and $v(x,y)$ are continuous. For small $\Delta x$ and $\Delta y$, we have
\begin{align*}
\Delta f&\approx df\\
&=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y\\
&=\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)\Delta x+\left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)\Delta y
\end{align*}
Dividing this by $\Delta z$, we obtain
\begin{align*}
\frac{\Delta f}{\Delta z}&\approx\frac{\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)\Delta x+\left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)\Delta y}{\Delta x+i\Delta y}\\
&=\frac{\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)+\left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)\frac{\Delta y}{\Delta x}}{1+i\frac{\Delta y}{\Delta x}}
\end{align*}
By the Cauchy-Riemann conditions, we have
$$\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}=i\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)$$
Therefore, the derivative $\frac{df}{dz}$ is given by
\begin{equation}
\label{eq:diff2}
\frac{df}{dz}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}
\end{equation}

Example. If $f(z)$ is differentiable at $z=z_0$, meaning $f'(z_0)$ exists and in some neighborhood of $z_0$, then we say that $f(z)$ is analytic at $z=z_0$.  If $f(z)$ is analytic everywhere in the complex plane $\mathbb{C}$, it is called an entire function.

Example. Let us consider $f(z)=z^2$. Let $z=x+iy$. Then $z^2= x^2-y^2+2xyi$, so $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$.
$$\frac{\partial u}{\partial x}=2x=\frac{\partial v}{\partial y}\ \mbox{and}\ \frac{\partial u}{\partial y}=-2y=-\frac{\partial v}{\partial x}$$
Hence, $f(z)=z^2$ satisfies the Cauchy-Riemann conditions. Since the partial derivatives are continuous, $f'(z)$ exists and
$$f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=2x+2yi$$

Example. Let $f(z)=\bar z=x-iy$. Then $u(x,y)=x$ and $v(x,y)=-y$. Since $\frac{\partial u}{\partial x}=1\ne -1=\frac{\partial v}{\partial y}$, $f(z)=\bar z$ is not differentiable at any $z=z_0$, and so it is nowhere analytic.

Remark. A function $f:\mathbb{C}\longrightarrow\mathbb{C}$ may be viewed as $f:\mathbb{R}^2\longrightarrow\mathbb{R}^2$, a vector-valued function of two real variables. However, the notion of differentiability is different between the two. For example, let us consider $f(z)=|z|^2$. It may be viewed as $f(x,y)=x^2+y^2$. As a real-valued function of two real variables, it is differentiable everywhere since $\frac{\partial f}{\partial x}=2x$ and $\frac{\partial f}{\partial y}=2y$ exist everywhere in $\mathbb{R}^2$. However, as a function of a complex variable, $u(x,y)=x^2+y^2$ and $v(x,y)=0$, so the Cauchy-Riemann conditions are not satisfied unless $x=y=0$. Hence, $f(z)=|z|^2$ is differentiable only at $(0,0)$. It is not analytic at $z=0$ nor anywhere else.