Cauchy-Riemann Equations
Theorem 1
Let the function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic at $\mathscr{R}$. If for the real function $u,v$ we have $$ f(z) = f(x+iy) = u(x,y) + iv(x,y) $$ then $u,v$ has a first-order partial derivative with respect to $x,y$ and satisfies the following system of partial differential equations at every point on $\mathscr{R}$. $$ \begin{cases} u_{x} (x,y) = v_{y} (x,y) \\ u_{y} (x,y) = -v_{x} (x,y) \end{cases} $$
Theorem
The Cauchy-Riemann equations can be summarized as follows.
$$ \begin{align*} f '(z) =& u_x + i v_x \\ =& v_y - i u_y \\ =& u_x -i u_y \\ =& v_y + i v_x \end{align*} $$
Polar Form 1
If $f \left( r e^{i \theta} \right) = u (r,\theta) + i v (r, \theta)$, then $$ \begin{cases} u_{r} (r, \theta) = {{ 1 } \over { r }} v_{\theta} (r,\theta) \\ v_{r} (r,\theta) = - {{ 1 } \over { r }} u_{\theta} (r,\theta) \end{cases} $$
Explanation
This theorem makes the otherwise unfamiliar concept of differentiation in the complex plane considerably more accessible. It demonstrates how similar the formulae for differentiation are between the real numbers and complex numbers, which is essential.
It is important to note that the converse is not always true. That is, even if the Cauchy-Riemann equations are satisfied $f$ may not be differentiable. The conditions for the converse to hold involve the continuity of partial derivatives among other things.
Proof
Since the function $f$ is differentiable at every point on $\mathscr{R}$, there exists a unique $f ' (z) = \lim_{h \to 0} {{f(z+h) - f(z)} \over {h}}$ regardless of the path $h \to 0$ taken.
Let $h=\alpha + i \beta$. If $\beta=0$ then $h$ will move along the real axis, and if $\alpha=0$ then $h$ will move along the imaginary axis. First, looking at the real axis, $$ \begin{align*} f '(z) =& \lim_{\alpha \to 0} {{ ( u(x+\alpha,y) - u(x,y) ) + i ( v(x+\alpha,y) - v(x,y) ) } \over {\alpha}} \\ =& u_{x} (x,y) + i v_{x} (x,y) \end{align*} $$ And examining the imaginary axis, $$ \begin{align*} f '(z) =& \lim_{\beta \to 0} {{ ( u(x,y+\beta) - u(x,y) ) + i ( v(x,y+\beta) - v(x,y) ) } \over {i \beta}} \\ =& { {u_{y} (x,y) + i v_{y} (x,y)} \over i } \\ =& v_{y} (x,y) - i u_{y} (x,y) \end{align*} $$ Note that this result stems from having $i \beta$ in the denominator when finding the limit.
Given that $f ' (z)$ is unique regardless of the path $h \to 0$ taken, by comparing the real and imaginary parts, it follows that $u_{x} (x,y) = v_{y} (x,y) $ and $-u_{y} (x,y) = v_{x} (x,y)$ must hold.
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