# Physical or geometric meaning of complex derivative [duplicate]

This question already has an answer here:

The derivative of a real-valued function at a point is the slope of the function at that point.

Similarly, what is the physical or geometric meaning of the derivative of a complex-valued function at a point?

## marked as duplicate by Claude Leibovici, Jonas Meyer, Paul Plummer, colormegone, user223391 Jul 2 '16 at 22:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Visual Complex Analysis by Tristan Needham might interest you. – ante.ceperic Aug 22 '14 at 14:32
• Really a interesting book. – Vrutang Aug 22 '14 at 18:12

## 3 Answers

You can think of the derivative as expressing the local/instantaneous stretching and rotation.

For example, suppose a real function $f$ has $f'(a)=2$, for some $a$. We can think of this as saying that near $a$ the function is (approximately) doubling distance, up to adding a constant. This is what the tangent line $2(x-a)+f(a)$ expresses.

Now for a complex function, $f'(a)=re^{i\theta}$ says that near $a$ we are (approximately) stretching/shrinking distances by the factor $r$, and also rotating by the angle $\theta$. For example $f'(a)=i$ means we are just rotating counterclockwise by $\pi/2$ radians. And $f'(a)=-i/2$ means we are shrinking by a factor of 2 and rotating counterclockwise by $3\pi/2$ radians.

One clear physical meaning is given by harmonic function theory. In particular, if $f=u+iv$ is complex-differentiable on an open set then we have that both the component functions satisfy Laplace's Equation. In other words, they can be thought of as possible potential energy functions for a force field which is conservative on the domain, or, as voltage functions for an electrostatic field on the domain. However, there is more, the relation between $u$ and $v$ is that they define level-curves which intersect at a right angle. That is, $u$ and $v$ are orthogonal functions. Hence, if we think of one as the potential energy then the other will play the role of the force. The tangent lines to the force field are perpendicular to the equipotential lines of the potential. All of this is wrapped up in the simple assumption that $f'(z)$ exist and is nonzero.

The function looks locally like $f(z)\approx f(z_0)+f'(z_0) (z-z_0)$. That's what it means.

Or you can phrase things in terms of linear algebra. I found these questions enlightening on the matter:

Why can the complex conjugate of a variable be treated as a constant when differentiating with respect to that variable?

Can we prove the Cauchy-Riemann equations using the matrix form of a complex number?

And my own question (with my own answer and no other input - what can I say): Is there a way to write the Cauchy-Riemann equation ∂f/∂z=0 without appealing to multivariable calculus?