The Cauchy-Riemann equations
Two partial differential equations tie the real part to the imaginary part: in a differentiable function they cannot be chosen separately. Harmonic functions follow from them and, with those, the calculation of fields.
01Two paths, one condition
In the previous topic it became clear that the complex derivative requires the difference quotient to give the same result from every direction. It is enough to impose that two directions agree to obtain a condition on the real and imaginary parts. Let us write \( f(z) = u(x,y) + j\,v(x,y) \) and approach first along the real axis, with \( \Delta z = \Delta x \):
And now along the imaginary axis, with \( \Delta z = j\,\Delta y \). Dividing by \( j \) is the same as multiplying by \( -j \):
The two expressions have to be the same complex number. Equating real part with real part and imaginary part with imaginary part gives the Cauchy-Riemann equations.
02The equations
That is the idea worth retaining: in a differentiable complex function, \( u \) and \( v \) are not independent. Once one is known, the other is determined up to a constant. It is what makes the complex case so different from the real one, where two functions of two variables owe nothing to each other.
Once they hold, the derivative can be computed with either of the two forms:
If \( z \) and \( \bar{z} \) are taken as independent variables, the two Cauchy-Riemann equations are equivalent to a single condition: \( \dfrac{\partial f}{\partial \bar{z}} = 0 \). A differentiable function has to depend on \( z \) alone. That is why \( \bar{z} \), \( |z|^2 \) and \( \operatorname{Re}(z) \) fail: all of them hide a \( \bar{z} \).
| Function | \( u,\ v \) | Satisfied? |
|---|---|---|
| \( z^2 \) | \( x^2 - y^2,\ 2xy \) | Yes: \( u_x = v_y = 2x \), \( u_y = -v_x = -2y \) |
| \( e^z \) | \( e^x\cos y,\ e^x \operatorname{sin} y \) | Yes, on the whole plane |
| \( \bar{z} \) | \( x,\ -y \) | No: \( u_x = 1 \) but \( v_y = -1 \) |
| \( |z|^2 \) | \( x^2 + y^2,\ 0 \) | Only at \( z = 0 \) |
| \( \operatorname{Re}(z) \) | \( x,\ 0 \) | No: \( u_x = 1 \ne v_y = 0 \) |
03The four derivatives, measured
The level curves of \( u \) and those of \( v \) say everything there is to know. If the function satisfies Cauchy-Riemann, the two families cross at right angles and the gradients have the same magnitude, because \( \nabla v \) is \( \nabla u \) rotated by 90°. Where that breaks down, the derivative does not exist.
Level curves of \( u \) in blue and of \( v \) in green, with the two gradients at the chosen point. Click or tap the canvas to move it.
04Necessary, and almost sufficient
Cauchy-Riemann is a necessary condition: if \( f \) is differentiable at \( z_0 \), the equations hold there. The converse is not enough. There are functions that satisfy both equations at a point and still are not differentiable, because the partial derivatives behave badly around it.
If \( u \) and \( v \) have continuous first-order partial derivatives in a neighborhood of \( z_0 \) and the Cauchy-Riemann equations hold there, then \( f \) is differentiable at \( z_0 \), and it is analytic on the whole neighborhood where that holds. In practice —polynomials, exponentials, rational functions away from their poles— continuity is guaranteed and checking the two equations is enough.
In polar coordinates, which is the convenient way to work with \( z^n \), \( \ln z \) and anything with circular symmetry, the equations become:
05Harmonic functions and the complex potential
Differentiating the first equation with respect to \( x \), the second with respect to \( y \) and adding, the cross terms cancel and what is left is:
Here the course crosses paths with Physics and with Fields. The electrostatic potential in a region with no charge satisfies exactly that equation. So, if \( u \) is taken as the potential, the function \( v \) that goes with it —its harmonic conjugate— has the field lines as its level curves, because they are orthogonal to the equipotentials. The two families seen in the lab are, in a real problem, the map of equipotentials and the map of lines of force.
Constructing the conjugate means integrating the equations. With \( u = x^2 - y^2 \):
Hence \( g'(x) = 0 \) and \( v = 2xy + C \): the function was \( f(z) = z^2 \) plus an imaginary constant. That procedure is what makes it possible, in a field problem, to obtain the lines of force from the potential without solving anything else.
The capacitance per unit length of a two-conductor line, the field at the edge of a microstrip and the characteristic impedance of a deformed coaxial cable are computed with this idea: the difficult geometry is transformed into a simple one with a conformal mapping, Laplace is solved there —where it is trivial— and you come back. Cauchy-Riemann is what guarantees that the problem stays the same after the transformation.
06In the lab
For \( f(z) = z^3 \), expand \( u \) and \( v \) in terms of \( x \) and \( y \), compute the four partial derivatives and check the two equations. Then obtain \( f'(z) \) with both forms from section 02 and verify that both give \( 3z^2 \).
Take \( f(z) = z\,\operatorname{Re}(z) = x^2 + jxy \) and find the points where the equations hold. Check that they hold only at the origin, and that therefore the function is not analytic at any point.
Set up an electrolytic tank with two electrodes and map, with a probe, the curves of equal potential on paper. Draw the orthogonal family on top by hand: these are the field lines, and they are the level curves of the harmonic conjugate of the potential.
07Common mistakes
- Getting the sign wrong. The second equation is \( u_y = -v_x \); writing it without the minus sign makes the conjugate come out analytic, which it is not.
- Believing that satisfying the equations is enough. Continuous partial derivatives on a neighborhood are also needed.
- Confusing a point with a neighborhood. \( |z|^2 \) satisfies Cauchy-Riemann at the origin and is not analytic anywhere.
- Choosing \( u \) and \( v \) separately. They are not independent: once one is fixed, the other is determined up to a constant.
- Using the Cartesian form where the polar form is better. With \( \ln z \) or \( z^{1/2} \), the polar version saves half the work.
08Self-assessment
Write the two equations without looking.
\( u_x = v_y \) and \( u_y = -v_x \). The minus sign goes in the second one.
Why doesn’t \( \bar{z} \) satisfy them?
With \( u = x \) and \( v = -y \): \( u_x = 1 \) while \( v_y = -1 \). The first one fails at every point of the plane.
If \( u = x^2 - y^2 \), what is its harmonic conjugate?
\( v = 2xy + C \), which reconstructs \( f(z) = z^2 \) up to a constant.
What geometric relationship is there between \( \nabla u \) and \( \nabla v \)?
They are perpendicular and of equal magnitude: \( \nabla v \) is \( \nabla u \) rotated by 90°. That is why the level curves cross at right angles.
What equation do \( u \) and \( v \) each satisfy on their own?
Laplace’s: \( u_{xx} + u_{yy} = 0 \). They are harmonic functions, which is the same equation that an electrostatic potential without charges satisfies.
Does satisfying the equations at a point guarantee that there is a derivative?
No. In addition, the partial derivatives must be continuous on a neighborhood; without that, there are counterexamples.
09Further reading
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The derivation of the equations, the sufficient condition and the polar form, with complete proofs.
- A. David Wunsch. Complex Variables with Applications. 2nd ed. (Spanish edition, “Variable compleja con aplicaciones”), Addison-Wesley Iberoamericana, 1997. The chapter on harmonic functions and the complex potential, designed for field problems.
- Matthew N. O. Sadiku. Elements of Electromagnetics. 5th ed. (Spanish edition, “Elementos de electromagnetismo”), Oxford University Press, 2018. Laplace in two dimensions, to see from the other side the same thing that is derived here from the complex derivative.