# Cauchy-Riemann Equations/Sufficient Condition

## Theorem

Let $D \subseteq \C$ be an open subset of the set of complex numbers $\C$.

Let $f: D \to \C$ be a complex function on $D$.

Let $u, v: \left\{ {\left({x, y}\right) \in \R^2: x + i y = z \in D }\right\} \to \R$ be two real-valued functions defined as:

$u \left({x, y}\right) = \operatorname{Re} \left({f \left({z}\right) }\right)$
$v \left({x, y}\right) = \operatorname{Im} \left({f \left({z}\right) }\right)$

where:

$\operatorname{Re} \left({f \left({z}\right)}\right)$ denotes the real part of $f \left({z}\right)$
$\operatorname{Im} \left({f \left({z}\right)}\right)$ denotes the imaginary part of $f \left({z}\right)$.

Let:

$u$ and $v$ be differentiable in their entire domain

and:

The following two equations, known as the Cauchy-Riemann equations, hold for the continuous partial derivatives of $u$ and $v$:
$(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$
$(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$

Then:

$f$ is complex-differentiable in $D$

and:

for all $z \in D$:
$f' \left({z}\right) = \dfrac {\partial f} {\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$

## Proof

Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire domain.

Let $h, k \in \R \setminus \left\{ {0}\right\}$, and put $t = h + i k \in \C$.

Let $\left({x, y}\right) \in \R^2$ be a point in the domain of $u$ and $v$.

Put:

$a = \dfrac{\partial u}{\partial x} \left({x, y}\right) = \dfrac{\partial v}{\partial y} \left({x, y}\right)$

and:

$b = - \dfrac{\partial u}{\partial y} \left({x, y}\right) = \dfrac{\partial v}{\partial x} \left({x, y}\right)$

From the Alternative Differentiability Condition, it follows that:

$u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \epsilon_{u x} \left({h}\right) }\right)$
$u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \epsilon_{u y} \left({k}\right) }\right)$
$v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \epsilon_{v x} \left({k}\right) }\right)$
$v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \epsilon_{v y} \left({h}\right) }\right)$

where $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are continuous real functions that converge to $0$ as $h$ and $k$ tend to $0$.

With $z = x + i y$, it follows that:

 $\displaystyle f \left({z + t}\right)$ $=$ $\displaystyle u \left({x + h, y + k}\right) + i v \left({x + h, y + k}\right)$ $\displaystyle$ $=$ $\displaystyle u \left({x, y}\right) + iv \left({x, y}\right) + h \left({a + i b + \epsilon_{u x} \left({h}\right) + i \epsilon_{v x} \left({h}\right)}\right) + k \left({i a - b + \epsilon_{u y} \left({k}\right) + i \epsilon_{v y} \left({k}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle f \left({z}\right) + t \left({a + i b + \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)}\right)$

With the assumption that the partial derivatives are continuous, we have $a + i b$,, which is a function of h, converges to itself by taking h -> 0.

Thus:

 $\displaystyle \epsilon \left({t}\right)$ $=$ $\displaystyle \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)$ $\displaystyle \implies \ \$ $\displaystyle \left\vert{\epsilon \left({t}\right) }\right\vert$ $\le$ $\displaystyle \left\vert{\dfrac h t \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{v y} \left({k}\right) }\right\vert$ Triangle Inequality for Complex Numbers $\displaystyle$ $=$ $\displaystyle \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert$ Complex Modulus of Product of Complex Numbers

Thus:

 $\displaystyle \lim_{t \mathop \to 0} \left\vert{\epsilon \left({t}\right) }\right\vert$ $\le$ $\displaystyle \lim_{t \mathop \to 0} \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert$ Combined Sum Rule for Limits of Functions $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert$ $\displaystyle$ $\le$ $\displaystyle \lim_{t \mathop \to 0} \left\vert{ \epsilon_{ux} \left({h}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vx} \left({h}\right) }\right\vert$ as $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by Modulus Larger than Real Part and Imaginary Part $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \lim_{t \mathop \to 0} \left\vert{ \epsilon_{uy} \left({k}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vy} \left({k}\right) }\right\vert$ $\displaystyle$ $=$ $\displaystyle 0$

This shows that:

$\displaystyle \lim_{t \mathop \to 0} \epsilon \left({t}\right) = 0$

From Continuity of Composite Mapping, it follows that $\epsilon \left({t}\right)$ is continuous.

Then the Alternative Differentiability Condition shows that:

$f' \left({z}\right) = a + i b$

$\Box$