27 Jul 2025
Functions of a Complex Variable, Analyticity, and Cauchy-Riemann Equations
Complex differentiation from path-independent limits, derivation of the Cauchy-Riemann equations, and analytic examples.
A function of a complex variable can be written
\[w=f(z)=u(x,y)+iv(x,y), \qquad z=x+iy,\]where $u$ and $v$ are real functions. Its derivative at $z_0$ is
\[\boxed{ f'(z_0)=\lim_{\Delta z\to0} \frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}}.\]Unlike a real increment, $\Delta z$ can approach zero from infinitely many directions. The same limit must be obtained along every path.
Derivation of the Cauchy-Riemann equations
First approach along the real direction: $\Delta z=\Delta x$. Then
\[f'(z)=\lim_{\Delta x\to0} \left(\frac{\Delta u}{\Delta x} +i\frac{\Delta v}{\Delta x}\right) =u_x+iv_x.\]Now approach along the imaginary direction: $\Delta z=i\Delta y$. Since $1/i=-i$,
\[\begin{aligned} f'(z) &=\lim_{\Delta y\to0} \frac{\Delta u+i\Delta v}{i\Delta y}\\ &=\lim_{\Delta y\to0} \left(\frac{\Delta v}{\Delta y} -i\frac{\Delta u}{\Delta y}\right)\\ &=v_y-iu_y. \end{aligned}\]Equality of real and imaginary parts gives
\[\boxed{u_x=v_y,\qquad u_y=-v_x}.\]These are the Cauchy-Riemann equations. When the first partial derivatives are continuous in a neighborhood, these equations are also sufficient for complex differentiability there.
To see the sufficiency, write
\[\Delta f=u_x\Delta x+u_y\Delta y +i(v_x\Delta x+v_y\Delta y)+o(\lvert\Delta z\rvert).\]Use $v_y=u_x$ and $v_x=-u_y$:
\[\begin{aligned} \Delta f &=(u_x-iu_y)(\Delta x+i\Delta y)+o(\lvert\Delta z\rvert)\\ &=(u_x+iv_x)\Delta z+o(\lvert\Delta z\rvert). \end{aligned}\]Division by $\Delta z$ and the limit $\Delta z\to0$ give
\[\boxed{f'(z)=u_x+iv_x=v_y-iu_y}.\]A function differentiable throughout an open neighborhood is analytic there.
Examples
For $f(z)=z^2$,
\[z^2=(x^2-y^2)+i(2xy),\]so
\[u_x=2x=v_y, \qquad u_y=-2y=-v_x.\]Thus $z^2$ is analytic everywhere and
\[f'(z)=2x+i2y=2z.\]
For $f(z)=\bar z=x-iy$,
\[u_x=1,\qquad v_y=-1,\]so the Cauchy-Riemann equations fail everywhere. Indeed,
\[\frac{\overline{\Delta z}}{\Delta z} =e^{-2i\arg(\Delta z)}\]depends on the direction of approach.
For $f(z)=\lvert z\rvert^2=x^2+y^2$, the Cauchy-Riemann equations hold only at the origin. The derivative there exists because
\[\frac{\lvert h\rvert^2}{h}=\bar h\longrightarrow0.\]However, the function is not analytic at the origin because it is not differentiable throughout any neighborhood of it. The Cauchy-Riemann residuals for the analytic and non-analytic examples are checked in the Unit III Maxima worksheet.
Discussion