📂Complex Anaylsis

Cauchy–Riemann Equations

Theorem¹

Let the function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic in $\mathscr{R}$. If, for real functions $u,v$, $$ f(z) = f(x+iy) = u(x,y) + iv(x,y) $$ then the first-order partial derivatives of $u,v$ with respect to $x,y$ exist and satisfy the following system of differential equations at every point of $\mathscr{R}$. $$ \begin{cases} u_{x} (x,y) = v_{y} (x,y) \\ u_{y} (x,y) = -v_{x} (x,y) \end{cases} $$

Summary

The Cauchy–Riemann equations are 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 Form¹

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 is a theorem that makes the unfamiliar notion of differentiation over the complex numbers considerably more approachable. It is absolutely essential for showing that the differentiation formulas over the real numbers and those over the complex numbers are similar.

Note that, in general, the converse does not hold. That is, even if the Cauchy–Riemann equations are satisfied, $f$ may fail to be differentiable. The condition under which the converse holds exists separately and is related to the continuity of the partial derivatives.

Proof

Since the function $f$ is differentiable at every point of $\mathscr{R}$, $f ' (z) = \lim_{h \to 0} {{f(z+h) - f(z)} \over {h}}$ exists uniquely regardless of the path along which $h \to 0$.

Let $h=\alpha + i \beta$. Here, if $\beta=0$ then $h$ moves along the real axis, and if $\alpha=0$ then $h$ moves 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 looking at 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 was obtained because $i \beta$ appears in the denominator when taking the limit.

Since the condition states that $f ' (z)$ is unique regardless of the path along which $h \to 0$, comparing the real and imaginary parts shows that $u_{x} (x,y) = v_{y} (x,y) $ and $-u_{y} (x,y) = v_{x} (x,y)$ must hold.

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