How do I plot complex functions using MATLAB?

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We begin with by defining a function on the complex domain z∈C$z\in \mathbb{C}$

g(z)=f(z)$g(z)=\mathit{\text{f}}(z)$

This is rather messy. 2 reasons why it’s hard to understand:

1. The complex numbers are not ordered. If we were badly to think of complex numbers as time (like we do for the real domain), which number would come first, 1$1$ or i$i$? Neither.
2. The input z$z$ has a pair of numbers to track. The output of the function has a pair of numbers to track. This makes 4$4$ numbers to track for any given z$z$. A plot that all at once would be visually 4D. Oof.

The best way to then plot it would be to extract what is important and discard what is not. Let us begin. Define a simpler, restricted domain such that z⊆C$z\subseteq \mathbb{C}$. One might choose the domain as the (bijective) unit circle z∈eix$z\in e^{ix}$ for 0≤x<2π|x∈R$0\le x<2\pi |x\in \mathbb{R}$, or something more exotic, like z∈eix+e2ix$z\in e^{ix}+e^{2ix}$, or a synthesized mesh, or a line, or a disc. Now in a way, our complex number’s are ordered. We visualize this domain on the z-plane. Then we simply apply our function and look at the range of the new resulting domain.

Define a morphology from the input domain to the output range, parametric by some linear weight k$k$ for 0≤k≤1$0\le k\le 1$.

Mk,z=g(z)k+z(1−k)$M_{k,z}=g(z)k+z(1-k)$

When k=0$k=0$, Mk,z=z$M_{k,z}=z$. As we slide to k=1$k=1$ our Mk,z$M_{k,z}$ morphs into g(z)$g(z)$. In this way, we can animate the path of our transformation (for visualization purposes only).

Example

A flat square with mass m$m$ sits restless at the center of a friction-less, flat square table. All 4 edges of the square are connected perpendicularly to ideal springs which are attached to rollers on the border of the table. Assume the springs are rigid in all but the direction of desired deflection. The west/east edges yield an equivalent spring constant ke,x$k_{e,x}$ and the north/south edges yield an equivalent spring constant ke,y$k_{e,y}$.

We pull the mass to a non-equilibrium position on the table, then release it with a known directional velocity. Determine any shape formed by the specified initial conditions of the spring. Use only transformations of the unit circle.

To solve this problem, we require only the simplest solvable ordinary differential equation known (thank goodness because it’s one of the only ones I can solve), the non-damped second order! Since the spring deflections are happening orthogonally, we treat the two directions as two separate problems.

−ke,xpx=md2pxdt2$-k_{e,x}{p}_x=m\frac{d^2{p}_x}{d{t}^2}$

−ke,ypy=md2pydt2$-k_{e,y}{p}_y=m\frac{d^2{p}_y}{d{t}^2}$

px(t)=Axcos(ke,x/m−−−−−−√t)+Bxsin(ke,x/m−−−−−−√t)$p_x(t)=A_{x}cos(\sqrt{k_{e,x}/m}t)+B_{x}sin(\sqrt{k_{e,x}/m}t)$

py(t)=Aycos(ke,y/m−−−−−−√t)+Bysin(ke,y/m−−−−−−√t)$p_y(t)=A_{y}cos(\sqrt{k_{e,y}/m}t)+B_{y}sin(\sqrt{k_{e,y}/m}t)$

Let the initial position and velocity be represented in vector form as <p0,x,p0,y>$<p_{0,x},p_{0,y}>$ and <v0,x,v0,y>$<v_{0,x},v_{0,y}>$ then solve for the constants Ax,y$A_{x,y}$ and Bx,y$B_{x,y}$.

Ax=p0,x;Bx=v0,xm/ke,x−−−−−−√$A_x=p_{0,x};B_x=v_{0,x}\sqrt{m/k_{e,x}}$

Ay=p0,y;By=v0,ym/ke,y−−−−−−√$A_y=p_{0,y};B_y=v_{0,y}\sqrt{m/k_{e,y}}$

We now use the complex plane to hold our solution.

p(t)=px(t)+ipy(t)$p(t)=p_x(t)+i{p}_y(t)$

When choosing z=eit$z=e^{it}$ as the unit circle we easily find that

p(z)=f(z)=AxRe(zke,x/m√)+BxIm(zke,x/m√)+i(AyRe(zke,y/m√)+ByIm(zke,y/m√))

Let’s get our head around this first. Mass doesn’t really matter. We have the specific case where ke,x=ke,y$k_{e,x}=k_{e,y}$ then we have the general case ke,x≠ke,y$k_{e,x}\ne k_{e,y}$. The non-equal case gets split into 2 types: when the resulting frequencies are never orthogonal (Type 1: dense), and when the frequencies are orthogonal (Type 2: periodic). If the frequencies are quickly orthogonal call them strongly periodic, if not call them weakly periodic. We plot these families of solutions over some long time interval.

Formally if ke,x=ke,y$k_{e,x}=k_{e,y}$

p(z)∈E$p(z)\in \mathbb{E}$

where E$\mathbb{E}$ designates the family of ellipses centered at zero defined on the complex plane. This includes all rotations and scalings of bounded lines, circles, and squished circles symmetric about a line passing through the origin.

If ke,x≠ke,y$k_{e,x}\ne k_{e,y}$, if we normalize our frequency set F=[wx,wy]/min(wx,wy)$F=[w_x,w_y]/min(w_x,w_y)$

and find that cF∉Z$cF\notin \mathbb{Z}$ for any c∈Z>0$c\in \mathbb{Z}>0$

Then we have determined a Type 1 - Dense solution. Let M=A2+B2−−−−−−−√$M=\sqrt{A^2+B^2}$ be amplitude magnitude and our output domain becomes the entire rectangle

p(z)∈C|−Mx≤Re(p(z))≤Mx;−iMy≤Im(p(z))i≤iMy$p(z)\in \mathbb{C}|-M_x\le Re(p(z))\le M_x;-i{M}_y\le Im(p(z))i\le i{M}_y$

If ke,x≠ke,y$k_{e,x}\ne k_{e,y}$, and we find that cF∈Z$cF\in \mathbb{Z}$ for any c∈Z>0$c\in \mathbb{Z}>0$ then we yield a closed periodic path, a Type 2 solution. Of course the following is true

p(z)⊆−Mx≤Re(p(z))≤Mx;−iMy≤Im(p(z))i≤iMy$p(z)\subseteq -M_x\le Re(p(z))\le M_x;-i{M}_y\le Im(p(z))i\le i{M}_y$

By periodicity

z∈eit$z\in e^{it}$ for t∈R|0≤t<2πcmin(wx,wy)