User:Anghel/Sandbox

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 $z \in D$:


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

Necessary condition
Let $z = x+iy \in D$.

The Alternative Differentiability Condition shows that there exists $r \in \R_{>0}$ such that for all $t \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:


 * $(\text i) \quad f\left({z + t}\right) = f \left({z}\right) + t \left({f' \left({z}\right) + \epsilon \left({t}\right) }\right)$

where $B_r \left({0}\right)$ denotes an open ball of $0$, and $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \mathbb C$ is a continuous function with $\displaystyle \lim_{t \to 0} \ \epsilon \left({t}\right) = 0$.

Define $a, b, h, k \in \R$ by $a+ib = f' \left({z}\right)$, and $h+ik = t$.

By taking the real parts of both sides of equation $(\text i)$, it follows that:

To find the partial derivative $\dfrac{\partial u}{\partial x}$, assume that $y$ is fixed, and let $t$ be wholly real.

Then $t = h$, and $k = 0$, so:


 * $u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \operatorname{Re}\left({ \epsilon \left({h}\right) }\right) }\right)$

From Limits of Real and Imaginary Parts, it follows that $\displaystyle \lim_{h \to 0} \operatorname{Re}\left({ \epsilon \left({h}\right) }\right) = \operatorname{Re}\left({ \lim_{h \to 0} \epsilon \left({h}\right) }\right)= \operatorname{Re}\left({0}\right) = 0$.

From Continuity of Composite Mapping and Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re}\left({ \epsilon \left({h}\right) }\right)$ is continuous as a function of $h$.

Then the Alternative Differentiability Condition proves that $\dfrac{\partial u}{\partial x} \left({x, y}\right) = a$.

To find the partial derivative $\dfrac{\partial u}{\partial y}$, assume that $x$ is fixed, and let $t$ be wholly imaginary.

Then $t = ik$, and $h = 0$, so:


 * $u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \operatorname{Re}\left({i \epsilon \left({ik}\right) }\right) }\right)$

From Limits of Real and Imaginary Parts, it follows that $\displaystyle \lim_{k \to 0} \operatorname{Re}\left({i \epsilon \left({k}\right) }\right) = \operatorname{Re}\left({i \lim_{k \to 0} \epsilon \left({k}\right) }\right) = 0$.

Then the Alternative Differentiability Condition proves that $\dfrac{\partial u}{\partial y} \left({x, y}\right) = -b$.

By taking the imaginary parts of both sides of equation $(\text i)$, it follows that:

To find the partial derivative $\dfrac{\partial v}{\partial x}$, assume that $y$ is fixed, and let $t$ be wholly real.

Then $t = h$, and $k = 0$, so:


 * $v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \operatorname{Im}\left({ \epsilon \left({h}\right) }\right) }\right)$

A similar argument to the ones above shows that the Alternative Differentiability Condition can be applied to prove that:


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

This proves the second Cauchy-Riemann equation.

To find the partial derivative $\dfrac{\partial v}{\partial y}$, assume that $x$ is fixed, and let $t$ be wholly imaginary.

Then $t = ik$, and $h = 0$, so:


 * $v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \operatorname{Im}\left({i \epsilon \left({ik}\right) }\right) }\right)$

Here, the Alternative Differentiability Condition shows that:


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

This proves the first Cauchy-Riemann equation.

From Complex-Differentiable Function is Analytic, it follows that $f'$ is continuous.

From Continuity of Composite Mapping and Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re} \left({f}\right)$ and $\operatorname{Im} \left({f}\right)$ are continuous.

Then all four partial derivatives are continuous, as


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


 * $\dfrac{\partial u}{\partial y} \left({x, y}\right) = - \dfrac{\partial v}{\partial x} \left({x, y}\right) = -b = -\operatorname{Im} \left({f \left({z}\right) }\right)$

By definition of differentiablity of real-valued functions, it follows that $u$ and $v$ are differentiable.

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:


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

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

Similarly: