Cauchy-Riemann Equations/Sufficient Condition

Theorem
Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is open.

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


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


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

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

Let:


 * $u$ and $v$ are differentiable in their entire domain


 * The following two equations, known as the Cauchy-Riemann equations, hold for the 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 $z \in D$:

P:$f' \left({z}\right) = \dfrac {\partial f} {\partial x} \left({z}\right) = -i \dfrac{\partial f}{\partial y} \left({z}\right)$

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:

With:
 * $\epsilon \left({t}\right) = \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)$

it follows that:

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:


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