Complex Numbers, Cauchy Riemann

Cauchy Riemann Condition

For simplicity, let f(0) = 0. Let f be differentiable at 0, with f′(0) = a+bi.

Let f have real component u and imaginary component v. In other words, f(x,y) = u(x,y) + v(x,y)i.

Since the difference quotient approaches its limit a+bi, it does so when h is restricted to the x and y axes. Let x approach 0 and consider the limit of f(x,0)/x. This gives the partials of u and v with respect to x, which must equal a and b respectively.

Next restrict h to the y axis, and evaluate f(0,y)/yi. This sets the partials of u and v (with respect to y) to -b and a. These four constraints form the Cauchy Riemann condition for complex differentiation.

   u∂x = a
   v∂x = b
   u∂y = -b
   v∂y = a

This condition is necessary for differentiability, and it becomes sufficient if the partials are continuous near 0. Let's borrow some theorems from multivariable calculus. Recall that u(x,y) is differentiable, as a two dimensional function, if its partials are continuous. The same holds for v(x,y). Now if u and v are both differentiable, then the composite vector function u,v is also differentiable.

What does this mean? For h near 0, f(h) is close to the jacobian (a,-b|b,a) multiplied by the vector h. But aha, this product is a+bi times h (when h is treated as a complex number). The error term goes to 0 relative to the length of h, and f(h)/h approaches a+bi. Thus f is differentiable at 0.

When the first partials are continuous about a point p, f is differentiable at p iff f satisfies the Cauchy Riemann condition.

Polar Coordinates

Let c be a point on the unit circle, whence z/c rotates the complex plane, pulling c back to 1 and ci back to i. The composite function f(z/c) is differentiable at 0, with derivative f′(0)/c. The Cauchy Riemann condition holds for this composite function. The partial of u with respect to x equals the partial of v with respect to y, and the partial of u with respect to y is minus the partial of v with respect to x. Spin the plane back, and the same relationships hold for the directional derivatives along the vectors c and ci. Conversely, if these relationships hold for the directional derivatives, and the partials are continuous, rotate the plane to show that f(z/c) is differentiable at 0. Compose with cz to show f(z) is differentiable.

We can now derive the Cauchy Riemann condition when u and v are given in polar coordinates. Given any point p in the complex plane, rotate the plane so that p lies on the x axis. By the above, f is differentiable at p iff the Cauchy Riemann condition holds. The partials with respect to x are the partials with respect to r, and the partials with respect to y are the partials with respect to θ divided by r. We can now derive the following criterion for differentiability.

r×u∂r = v∂θ
r×v∂r = -u∂θ

Derivative Equals Zero

Let's use the Cauchy Riemann condition to show an analytic function f with derivative 0 is constant. The partials of u, and the partials of v, are everywhere 0. Suppose f(p) ≠ f(q). Without loss of generality, suppose u(p) ≠ u(q). Move p parallel to the x axis, until it is directly above or below q in the complex plane. Now u cannot change, since the partial of u with respect to x is always 0. Then move p up to q, and again, u cannot change, since the partial of u with respect to y is always 0. Therefore f is constant.

Bicontinuous Function

Assume f is continuously differentiable throughout a region, and f′ is never 0. View f as a real function from R² into R². The jacobian has a determinant of a²+b², which is nonzero. This implies open sets map to open sets, and f is bicontinuous. If f happens to be injective from one region onto another, it implements a homeomorphism between the two regions.

Later on we will see that differentiable implies continuously differentiable, and the zero derivatives of f are isolated points, and this allows us to generalize the theorem. If f is nonconstant and differentiable throughout a region it is bicontinuous. If f has a zero derivative at p, f+z is bicontinuous about p, and when we subtract the bicontinuous function z, f is bicontinuous about p.