We have seen that, among vectors, the curl of a function of r times a unit vector in the r dimension is 0 where the derivative is defined. There is a similar statement about any function, f of the variable z, with z = x + iy.
We can deduce it, by applying the chain rule for differentiating f(z)
Since is 1, and is i, we find
and upon taking real and imaginary parts of this equation we get
These conditions tell us that the "gradient vectors" of the real
and imaginary parts of a function of x + iy are orthogonal everywhere in the
Combining these two equations tells us that the two dimensional Laplacian of either the real or imaginary part of any twice differentiable function of z is 0.
Exercise 18.1 Deduce this property explicitly here.
We can represent either the real part or the imaginary part of any function f(z) as a real function of the two variables x and y.
We can do so as a three dimensional image, or can use the "equipotential" approach of representing the either function by contour lines in a planar figure.
We noticed when discussing that representation that we could also connect the gradient vectors at appropriate points together into paths, which are sometimes called "lines of force" as they are in certain physical contexts.
What the relations we have been discussing tell us is that the contour lines for the real part of any suitably differentiable function of z are lines of force for the imaginary part and vice versa!
For example, consider the function z2. Its real part is x2 - y2 and its imaginary part is 2xy.
18.2 Look at the contour lines for each using the applet; and verify that the contour lines for one are normal to those for the other.
18.3 Try it for for x and y both less than 1.