Question 2.3.5: A Linear Mapping of a Triangle. Find a complex linear functi......

Let S_{1} denote the triangle with vertices 1 + i, 2 + i, and \frac{3}{2} + (1 +\frac{ 1}{2}\sqrt{3})i shown in color in Figure 2.15(a), and let S^{\prime} represent the triangle with vertices i, 3i, and \sqrt{3} + 2i shown in black in Figure 2.15(d). There are many ways to find a linear mapping that maps S_{1} onto S^{\prime }. One approach is the following: We first translate S_{1} to have one of its vertices at the origin. If we decide that the vertex 1 + i should be mapped onto 0, then this is accomplished by the translation T_{1}(z) = z − (1 + i). Let S_{2} be the image of S_{1} under T_{1}. Then S_{2} is the triangle with vertices 0, 1, and \frac{1}{2} + \frac{1}{2}\sqrt{3} i shown in black in Figure 2.15(a). From Figure 2.15(a), we see that the angle between the  imaginary axis and the edge of S_{2} containing the vertices 0 and \frac{1}{2} + \frac{1}{2}\sqrt{3} i is π/6. Thus, a rotation through an angle of π/6 radians counterclockwise about the origin will map S_{2} onto a triangle with two vertices on the imaginary axis. This rotation is given by R(z) =(e^{iπ/6}) z = (\frac{1}{2}\sqrt{3} +\frac{ 1}{2} i)z, and the image of S_{2} under R is the triangle S_{3} with vertices at 0,\frac{ 1}{2}\sqrt{3} +\frac{ 1}{2} i, and i shown in black in Figure 2.15(b). It is easy to verify that each side of the triangle S_{3} has length 1. Because each side of the desired triangle S^{\prime } has length 2, we next magnify S_{3} by a factor of 2. The magnification M(z) = 2z maps the triangle S_{3} shown in color in Figure 2.15(c) onto the triangle S_{4} with vertices 0,\sqrt{3} + i, and 2i shown in black in Figure 2.15(c). Finally, we translate S_{4} by i using the mapping T_{2}(z) = z +i. This translation maps the triangle S_{4} shown in color in Figure 2.15(d) onto the triangle S^{\prime} with vertices i,\sqrt{3} + 2i, and 3i shown in black in Figure 2.15(d).