Visualizing complex functions
In this set of class notes, we’re going to take a look at an important way to visualize complex functions to, hopefully, generate some intuition about how they operate.
Why not graphs??
I guess the most basic way to visualize a function is via a graph. We’re certainly all familiar with the graph of \(f(x)=x^2\):
Note that this function, like most of the functions that you study in calculus, maps \(\mathbb R \to \mathbb R\). We say that the point \((x,y)\) is on the graph of \(y=f(x)\). Thus, the graph lives in \(\mathbb R^2\).
Now, if \(f:\mathbb{C}\to\mathbb{C}\), then perhaps its graph lives in \(\mathbb{C}^2\). Unfortunately, each copy of \(\mathbb{C}\) requires two real dimensions to draw. Thus, our image would look four dimensional.
I guess we’ll need another approach.
Lineland
In search of an alternative, let’s suppose we live in one-dimensional space (called lineland in Edwin Abbott’s classic flatland) and wish to visualize a function \(f:\mathbb R \to \mathbb R\). One way to do so might be to illustrate the effect of \(f\) by viewing before and after images of natural sets in the line. We might even do so via an animation, like so:
Back here in our universe, we might realize that what we are effectively doing is examining the image of the set on the \(x\)-axis after it’s been mapped to the \(y\)-axis.
Complex functions
The analogous approach for complex functions goes like so:
- Pick a domain \(D\subset\mathbb{C}\), then
- Plot \(\{f(z)\in\mathbb{C}: z\in D\}\).
Often, it helps to shade and place a mesh on \(D\) to help see more precisely what maps to what.
The effect of \(z^2\) on a square
The image below shows the complex square \[\{z\in\mathbb{C}: 0\leq \text{Re}(z) \leq 2, \: 0 \leq \text{Im}(z) \leq 2\}.\] You can hit the play button to see the effect of the complex square function \(f(z)=z^2\) on this square - or, actually, the effect of \(f_p(z) = z^p\), as \(t\) ranges from \(p=1\) to \(p=2\).
To understand this, it helps to consider the images of the sides. For example, you might consider the right side \[S = \{2+t\,i: 0\leq t \leq 2\}.\] Note that \[(2+t\,i)^2 = (4-t^2) + 4t\,i.\] As \(t\) ranges from 0 to 2, this traces out a parabolic arc that moves from \(4\) to \(8i\). That’s exactly the right side that you see after the deformation.
The exponential function
Here’s a similar animation for the exponential function applied to the unit square \[\{z\in\mathbb{C}: 0\leq \text{Re}(z) \leq 1, \: 0 \leq \text{Im}(z) \leq 1\}.\] Now, we see the unit square morph into its image under the exponential \(z\to e^z\).
It again helps to consider how each side is determined.
The reciprocal function
This example shows how the image of the unit square morphs under application of \(f_p(z) = t^p\), as \(p\) ranges from \(1\) down to \(-1\). The final result should be the image of the unit square under the application of the reciprocal.
The straight sides should make sense. The others might be tricky.
Polar domains
In principle, you can try this with any input domain that you want. I’ve only programmed it for both Cartesian rectangles and polar wedges, though. The latter has the form \[\{re^{i\theta}: a\leq r \leq b \text{ and } \alpha\leq\theta\leq\beta\}.\]
Power functions
In the image below, \(a=1/2\), \(b=2\), \(\alpha=0\), and \(\beta=\pi/2\). The slider controls the value of \(p\) in \(f_p(z) = z^p\) and that value ranges from \(1\) to \(2\). Thus, the image morphs from the initial wedge to its image under the square function as \(p\) ranges from \(1\) to \(2\)
This one seems eminently easy to reason with because the action of a power function interacts so naturally with the shape of the region. We have \[r \, e^{i\theta} \to r^2 e^{2i\theta}.\] Thus, the distance of any point to the origin is squared and its argument doubles.
A Mobius transformation
We’ll have fun with these. This one shows the image of the unit circle under the function \[f(z) = i\frac{1-z}{1+z}.\]
Not so nice
Things can get confusing when the function is not one-to-one on its domain.