Theorem: let f(z) = u(x,y) + iv(x,y) be a single-valued complex function defined in a domain D \subseteq \mathbb{C}. Suppose that the partial derivatives \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, and \frac{\partial v}{\partial y} exist and are continuous at every point z in D. Then, f(z) is analytic in D if and only if the Cauchy-Riemann conditions are satisfied at every point z in D.
The Cauchy-Riemann conditions are:
\begin{aligned} & \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \\ & \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} \end{aligned}
This theorem has two parts: the necessity of the Cauchy-Riemann conditions for analyticity, and their sufficiency when coupled with the continuity of the partial derivatives.
Assumption: The function f(z) is analytic in the domain D. This means that for any point z_0 = x_0 + iy_0 in D, the complex derivative f^\prime(z_0) exists.
The definition of the complex derivative is:
f^\prime(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z}
Since f(z) is analytic, this limit must exist and be unique, regardless of the path along which \Delta z approaches 0. Let \Delta z = \Delta x + i\Delta y.
Considering path 1 where \Delta z approaches 0 along the real axis (\Delta y = 0, so \Delta z = \Delta x):
\begin{aligned} f^\prime(z_0) & = \lim_{\Delta x \to 0} \frac{u(x_0 + \Delta x, y_0) + iv(x_0 + \Delta x, y_0) - (u(x_0, y_0) + iv(x_0, y_0))}{\Delta x} \\ & = \lim_{\Delta x \to 0} \left[ \frac{u(x_0 + \Delta x, y_0) - u(x_0, y_0)}{\Delta x} + i \frac{v(x_0 + \Delta x, y_0) - v(x_0, y_0)}{\Delta x} \right] \end{aligned}
By the definition of partial derivatives, this becomes:
f^\prime(z_0) = \frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)
Considering path 2 where \Delta z approaches 0 along the imaginary axis (\Delta x = 0, so \Delta z = i\Delta y):
\begin{aligned} f^\prime(z_0) & = \lim_{\Delta y \to 0} \frac{u(x_0, y_0 + \Delta y) + iv(x_0, y_0 + \Delta y) - (u(x_0, y_0) + iv(x_0, y_0))}{i\Delta y} \\ & = \lim_{\Delta y \to 0} \left[ \frac{u(x_0, y_0 + \Delta y) - u(x_0, y_0)}{i\Delta y} + i \frac{v(x_0, y_0 + \Delta y) - v(x_0, y_0)}{i\Delta y} \right] \\ & = \lim_{\Delta y \to 0} \left[ \frac{1}{i} \frac{u(x_0, y_0 + \Delta y) - u(x_0, y_0)}{\Delta y} + \frac{v(x_0, y_0 + \Delta y) - v(x_0, y_0)}{\Delta y} \right] \end{aligned}
Since \frac{1}{i} = -i, this becomes:
f^\prime(z_0) = -i \frac{\partial u}{\partial y}(x_0, y_0) + \frac{\partial v}{\partial y}(x_0, y_0)
Rearranging the terms:
f^\prime(z_0) = \frac{\partial v}{\partial y}(x_0, y_0) - i \frac{\partial u}{\partial y}(x_0, y_0)
Since f^\prime(z_0) must be unique, we equate the real and imaginary parts of the expressions:
\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y}
Equating the real parts:
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}
Equating the imaginary parts:
\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}
which is equivalent to:
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
These are the Cauchy-Riemann conditions.
If f(z) is analytic in D, the Cauchy-Riemann conditions must be satisfied at every point in D.
Assumptions: the Cauchy-Riemann conditions hold at a point z_0 = x_0 + iy_0 in D:
\begin{aligned} & \frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0) \\ & \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0) \end{aligned}
Furthermore, partial derivatives \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y} are continuous in a neighborhood of (x_0, y_0).
We want to show that exists f^\prime(z_0).
Let:
\begin{aligned} & \Delta z = \Delta x + i\Delta y \\ & f(z_0 + \Delta z) - f(z_0) = \Delta f = \Delta u + i\Delta v \\ & \Delta u = u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0) \\ & \Delta v = v(x_0 + \Delta x, y_0 + \Delta y) - v(x_0, y_0) \end{aligned}
Since the partial derivatives of u and v are continuous, u(x,y) and v(x,y) are differentiable (in the real sense) at (x_0, y_0), we can write the total derivatives:
\begin{aligned} \Delta u &= \frac{\partial u}{\partial x} \Delta x + \frac{\partial u}{\partial y} \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y \\ \Delta v &= \frac{\partial v}{\partial x} \Delta x + \frac{\partial v}{\partial y} \Delta y + \eta_1 \Delta x + \eta_2 \Delta y \end{aligned}
where all partial derivatives are evaluated at (x_0, y_0), and \epsilon_1, \epsilon_2, \eta_1, \eta_2 \to 0 as (\Delta x, \Delta y) \to (0,0).
So:
\Delta f = \left( \frac{\partial u}{\partial x} \Delta x + \frac{\partial u}{\partial y} \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y \right) + i \left( \frac{\partial v}{\partial x} \Delta x + \frac{\partial v}{\partial y} \Delta y + \eta_1 \Delta x + \eta_2 \Delta y \right)
Now, using the Cauchy-Riemann conditions and substituting:
\Delta f = \left( \frac{\partial u}{\partial x} \Delta x - \frac{\partial v}{\partial x} \Delta y + \epsilon_1 \Delta x + \epsilon_2 \Delta y \right) + i \left( \frac{\partial v}{\partial x} \Delta x + \frac{\partial u}{\partial x} \Delta y + \eta_1 \Delta x + \eta_2 \Delta y \right)
Rearranging terms:
\begin{aligned} \Delta f & = \frac{\partial u}{\partial x} (\Delta x + i\Delta y) + i \frac{\partial v}{\partial x} (i\Delta y + \Delta x) + (\epsilon_1 \Delta x + \epsilon_2 \Delta y) + i(\eta_1 \Delta x + \eta_2 \Delta y) \\ & = \frac{\partial u}{\partial x} \Delta z + i \frac{\partial v}{\partial x} \Delta z + (\epsilon_1 + i\eta_1)\Delta x + (\epsilon_2 + i\eta_2)\Delta y \end{aligned}
Dividing by \Delta z = \Delta x + i\Delta y:
\frac{\Delta f}{\Delta z} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} + (\epsilon_1 + i\eta_1)\frac{\Delta x}{\Delta z} + (\epsilon_2 + i\eta_2)\frac{\Delta y}{\Delta z}
We need to show that the last two terms go to zero as \Delta z \to 0.
Since:
\begin{aligned} & |\Delta x| \le |\Delta z| \\ & |\Delta y| \le |\Delta z| \end{aligned}
we have:
\begin{aligned} & \left|\frac{\Delta x}{\Delta z}\right| \le 1\\ & \left|\frac{\Delta y}{\Delta z}\right| \le 1 \end{aligned}
As \Delta z \to 0 we have that:
\begin{aligned} & \Delta x \to 0\\ & \Delta y \to 0 \end{aligned}
and therefore:
\epsilon_1, \epsilon_2, \eta_1, \eta_2 \to 0 So:
\lim_{\Delta z \to 0} \left[ (\epsilon_1 + i\eta_1)\frac{\Delta x}{\Delta z} + (\epsilon_2 + i\eta_2)\frac{\Delta y}{\Delta z} \right] = 0
Removing these terms gives:
\lim_{\Delta z \to 0} \frac{\Delta f}{\Delta z} = \frac{\partial u}{\partial x}(x_0, y_0) + i \frac{\partial v}{\partial x}(x_0, y_0)
This limit exists, so f(z) is differentiable at z_0, and its derivative is:
f^\prime(z_0) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}
Using the Cauchy-Riemann conditions, we can also write:
f^\prime(z_0) = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y}
Since z_0 was an arbitrary point in D where the Cauchy-Riemann conditions hold and the partial derivatives are continuous, f(z) is analytic in D. This completes the proof of sufficiency.