Since both the domain and the range of such a function have dimension two -- as a real space -- the graph will be an object in real 4-space. Thus, it cannot be represented directly. We will have to use other devices.

If you have not already done so in Part 12, enter the following line to load the plots package:
with(plots);

1. We begin by thinking of the function f(z) = z² as a map of one copy of the complex plane into another. Enter the following commands to see where the square function maps a short vertical line segment. Click on the 1:1 button to obtain the proper aspect ratio. Experiment with values of p between -1 and 1.
   p := -0.5;
   complexplot((p + t*I)^2, t = -1..1, view=[-1..1,-2..2]);

2. Alter the code to see how the function maps short horizontal segments.

3. Maple has a graphics command that shows what happens to a mesh of short horizontal and vertical line segments under the function. First, enter the following to see the line segments themselves. Again, click on the 1:1 button to obtain the proper aspect ratio. (This should be done at each step for the remainder of this part.)
   conformal(z, z =-1 - I..1 + I);
   Now change the first occurrence of z in the command to z² to see the images under the square function, i.e., enter
   conformal(z^2, z =-1 - I..1 + I);
   See if you can identify where each line segment is mapped. Why are there fewer curves than there were line segments?

4. Another representation option is to reduce the number of dimensions to three by plotting |f(z)| against z. The command complexplot3d does this and colors the plot to indicate changes in the argument of f(z). Enter the following:
   complexplot3d(z^2, z=-1 - I..1 + I);

5. For still another representation, we'll concentrate on how the square function maps pieces of the unit circle. Enter the following:
   theta := Pi/4;
   complexplot((cos(t) + sin(t) * I)^2, t = 0 .. theta, view = [-1..1,-1..1]);
   Now increase theta. As you see, eventually the image starts to draw over itself. To indicate more clearly what is going on, we'll increase the vertical component as the curve is drawn out. Enter the following:
   g := t -> cos(t) + sin(t)*I;
   p := 3*Pi/2;
   spacecurve([Re(g(t)^2), Im(g(t)^2), t], t = 0 .. p, numpoints = 200);
   Now increase p from 3Pi/2 to 2Pi.

6. Give a word description of the square function.

7. Use your description to explain why every nonzero complex number has exactly two square roots. Give a geometric method of finding them, i.e., if you have located a non-zero complex number in the plane, where should you look for the square roots?

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