Complex Differentiation and the Cauchy–Riemann Equations

In this lesson you’ll derive the Cauchy–Riemann equations from the requirement that the derivative be direction-independent, learn exactly when they are sufficient as well as necessary, and use them as a fast test for differentiability.

$f$ is differentiable at $z_0$ if

$$f'(z_0) = \lim_{h\to0}\frac{f(z_0+h)-f(z_0)}{h}$$

exists, with $h$ complex and the limit independent of how $h\to0$.

The differentiation rules you know all hold with the same proofs: sum, product, quotient, chain, and $\frac{d}{dz}z^n = nz^{n-1}$. Differentiability implies continuity, again with the same argument.

What is new is how few functions qualify.

Write $f = u+iv$ and let $h$ approach 0 in two particular ways.

Along the real axis, $h = t$ with $t$ real:

$$f'(z_0) = \lim_{t\to0}\frac{u(x_0+t,y_0)-u(x_0,y_0) + i\big[v(x_0+t,y_0)-v(x_0,y_0)\big]}{t} = u_x + iv_x$$

Along the imaginary axis, $h = it$:

$$f'(z_0) = \lim_{t\to0}\frac{u(x_0,y_0+t)-u(x_0,y_0)+i\big[v(x_0,y_0+t)-v(x_0,y_0)\big]}{it} = \frac{u_y+iv_y}{i} = v_y - iu_y$$

If $f$ is differentiable these two must be equal. Matching real and imaginary parts gives the Cauchy–Riemann equations:

$$\boxed{\;u_x = v_y \qquad\text{and}\qquad u_y = -v_x\;}$$

And the derivative itself has two equivalent formulas:

$$f'(z) = u_x + iv_x = v_y - iu_y$$

Note that we used only two directions. Every other direction imposes no new condition - that is a small miracle and the reason the theory is workable at all.

The derivation shows the equations are necessary. On their own they are not quite sufficient, and the standard counterexample is unpleasant enough that it is worth knowing the correct statement instead:

Sufficiency. If $u, v$ have continuous first partial derivatives on a neighbourhood of $z_0$ and satisfy the Cauchy–Riemann equations there, then $f$ is differentiable at $z_0$.

In practice this is what you use. Both directions are needed for different purposes: necessity lets you rule functions out with a one-line check, sufficiency lets you rule them in.

Three readings, all worth carrying.

Rigidity. The equations couple $u$ and $v$ tightly: you cannot choose the real and imaginary parts independently. Given $u$, the function $v$ is determined up to a constant. Half of an analytic function determines the other half.

No conjugates allowed. Introduce the operator

$$\frac{\partial f}{\partial\bar z} = \frac12\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)$$

The Cauchy–Riemann equations are exactly $\frac{\partial f}{\partial\bar z} = 0$, which says $f$ depends on $z$ but not on $\bar z$. That is the sharpest way to state the whole condition, and it instantly explains why $\bar z$, $|z|^2 = z\bar z$, $\operatorname{Re}z = \frac{z+\bar z}{2}$, and $\operatorname{Im}z$ all fail.

Geometric. As a real map $\mathbb{R}^2\to\mathbb{R}^2$, the derivative of $f$ has Jacobian

$$\begin{pmatrix} u_x & u_y \\ v_x & v_y\end{pmatrix} = \begin{pmatrix} a & -b \\ b & a\end{pmatrix}$$

after substituting the equations. Matrices of that shape are exactly the rotation-and-scaling matrices. So complex differentiability means the map is locally a rotation and a uniform scaling, which is why analytic maps preserve angles. Conjugation’s Jacobian is a reflection, which is why it fails.

Also worth noting: the Jacobian determinant is $a^2+b^2 = |f'(z)|^2$, so $|f'|^2$ is the local area magnification factor.

Differentiate the first equation with respect to $x$ and the second with respect to $y$, then add. Assuming enough smoothness, which we will later prove is automatic,

$$u_{xx}+u_{yy} = 0$$

and the same for $v$. So the real and imaginary parts of an analytic function are harmonic: they satisfy Laplace’s equation. This is the bridge to physics, since Laplace’s equation governs steady-state heat, electrostatic potential, and incompressible irrotational flow. It gets a lesson of its own later.

For functions naturally written in polar coordinates, with $f = u(r,\theta)+iv(r,\theta)$:

$$\frac{\partial u}{\partial r} = \frac1r\frac{\partial v}{\partial\theta}, \qquad \frac{\partial v}{\partial r} = -\frac1r\frac{\partial u}{\partial\theta}$$

These are what you use for $\log z$ and for power functions with non-integer exponents.

Example 1: Verify $f(z)=z^2$ is differentiable and find $f'$.

Solution. Here $u = x^2-y^2$ and $v = 2xy$, so

$$u_x = 2x, \quad v_y = 2x, \quad u_y = -2y, \quad v_x = 2y$$

Both equations hold: $u_x = v_y$ ✓ and $u_y = -v_x$ ✓. The partials are polynomials, hence continuous, so $f$ is differentiable everywhere.

$$f' = u_x+iv_x = 2x+2iy = 2z$$

✓ as expected.

Example 2: Show $f(z)=\bar z$ fails the test.

Solution. $u = x$ and $v = -y$, giving

$$u_x = 1, \quad v_y = -1, \quad u_y = 0, \quad v_x = 0$$

The second equation holds, since $0 = -0$. The first does not: $1 \ne -1$, and not at any point.

Answer: nowhere differentiable. ∎

One failed equation is enough, and here it fails by a constant amount everywhere, so there is not even a single exceptional point.

Example 3: Where is $f(z) = |z|^2$ differentiable?

Solution. $|z|^2 = x^2+y^2$, so $u = x^2+y^2$ and $v = 0$.

$$u_x = 2x, \quad v_y = 0, \quad u_y = 2y, \quad v_x = 0$$

The equations require $2x = 0$ and $2y = 0$, so both hold only at the origin. The partials are continuous, so $f$ is differentiable at $z=0$ and nowhere else, with $f'(0) = 0$.

Differentiable at a single point is useless, because every strong theorem in this subject needs differentiability on an open set. This is exactly why the word analytic is defined the way it is in the next lesson.

Example 4: Verify $f(z)=e^z$ is entire.

Solution. $e^z = e^x\cos y + ie^x\sin y$, so $u = e^x\cos y$ and $v = e^x\sin y$. Then

$$u_x = e^x\cos y, \quad v_y = e^x\cos y, \quad u_y = -e^x\sin y, \quad v_x = e^x\sin y$$

Both equations hold everywhere, and all partials are continuous. So $e^z$ is differentiable on all of $\mathbb{C}$, and

$$f' = u_x+iv_x = e^x\cos y + ie^x\sin y = e^z$$

The exponential is its own derivative in $\mathbb{C}$ too. ∎

Example 5: Find a harmonic conjugate.

Given $u(x,y) = x^2-y^2+3x$, find $v$ so that $u+iv$ is analytic.

Solution. From the first equation, $v_y = u_x = 2x+3$, so integrating in $y$,

$$v = 2xy+3y+g(x)$$

for some function $g$ of $x$ alone. From the second equation, $v_x = -u_y = 2y$. But differentiating the expression above gives $v_x = 2y+g'(x)$, so $g'(x)=0$ and $g$ is a constant $C$.

$$v = 2xy+3y+C$$

Checking, $f = z^2+3z + iC$. ✓

The constant is the only freedom, which is the rigidity from earlier made concrete: $u$ determines $v$ up to an additive constant.

Example 6: A function satisfying the equations at a point yet not differentiable there.

Solution. Let $f(z) = \frac{\bar z^2}{z}$ for $z \ne 0$ and $f(0)=0$. A computation shows $u$ and $v$ satisfy the Cauchy–Riemann equations at the origin, because all four partials there are 0.

But along the ray $\arg z = \theta$,

$$\frac{f(z)-f(0)}{z} = \frac{\bar z^2}{z^2} = e^{-4i\theta}$$

which depends on $\theta$, so the limit does not exist and $f$ is not differentiable at 0.

This is why the sufficiency statement requires the partials to be continuous in a neighbourhood, not merely to satisfy the equations at the one point. The partials here are not continuous at 0.

Example 7: Use the polar equations on $\log z$.

Solution. Take the principal branch $\log z = \ln r + i\theta$, so $u = \ln r$ and $v = \theta$. Then

$$u_r = \frac1r, \qquad \frac1r v_\theta = \frac1r \cdot 1 = \frac1r \quad ✓$$ $$v_r = 0, \qquad -\frac1r u_\theta = 0 \quad ✓$$

Both hold for $r>0$, so $\log z$ is differentiable away from the origin and away from whatever ray we cut to make $\theta$ single-valued. And

$$\frac{d}{dz}\log z = \frac1z$$

∎

Fluid flow. For a two-dimensional incompressible irrotational flow, the velocity potential and stream function are precisely a pair $u, v$ satisfying the Cauchy–Riemann equations. The complex function $u+iv$ is the complex potential, and its derivative gives the velocity field directly.

Electrostatics. Potential and field lines form the same orthogonal pair. Field-line diagrams in textbooks are level curves of $u$ and $v$ for an analytic function, which is why they always cross at right angles.

Conformal meshing. Grid generation for finite-element simulation often uses analytic maps because the Jacobian being a rotation-and-scaling means cells stay well shaped, which keeps the numerics stable.

Image processing. The Wirtinger derivative $\partial/\partial\bar z$ is used in complex-valued signal and image processing, where a filter’s dependence on $\bar z$ measures how far it is from being analytic.

Elasticity. The Kolosov–Muskhelishvili approach solves plane elasticity problems using two analytic functions, turning a system of PDEs into complex function theory. It is standard in fracture mechanics.

Aerodynamics. Circulation and lift are computed from a complex potential whose real and imaginary parts satisfy these equations, which is how the Kutta–Joukowski theorem is derived.