Cauchy-Riemann Equations/Sufficient Condition

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

Let $u, v: \left\{ {\left({x, y}\right) \in \R^2 }\, \middle\vert \, {x+iy = z \in D }\right\} \to \R$ 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 $f \left({z}\right)$.

Then $f$ is complex-differentiable in $D$ iff


 * $u$ 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 $v$:


 * $(1): \quad \dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$
 * $(2): \quad \dfrac{\partial u}{\partial y} = - \dfrac{\partial v}{\partial x}$

If the conditions are true, then for all $z \in D$:


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

Sufficient condition
Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire Definition:Domain.

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

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:


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

where $\epsilon_{ux}, \epsilon_{uy}, \epsilon_{vx}, \epsilon_{vy}$ are continuous real functions that converge to $0$ as $h$ and $k$ tend to $0$.

With $z = x+iy$, it follows that:

With $\epsilon \left({t}\right) = \dfrac h t \epsilon_{ux} \left({h}\right) + \dfrac h t i \epsilon_{vx} \left({h}\right) - \dfrac k t i \epsilon_{uy} \left({k}\right) + \dfrac k t \epsilon_{vy} \left({k}\right)$, it follows that:

This shows that $\displaystyle \lim_{t \to 0} \epsilon \left({t}\right) = 0$.

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

Then the Alternative Differentiability Condition shows that:


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