Question 2.4.3: Image of a Triangle under w = z². Find the image of the tr......

Let S denote the triangle with vertices at 0, 1 + i, and 1 − i, and let S^{\prime} denote its image under w = z². Each of the three sides of S will be treated separately. The side of S containing the vertices 0 and 1 + i lies on a raye manating from the origin and making an angle of π/4 radians with the positive x-axis. By our previous discussion, the image of this segment must lie on a raymaking an angle of 2 (π/4) = π/2 radians with the positive u-axis. Furthermore, since the moduli of the points on the edge containing 0 and 1+i vary from 0 to \sqrt{2} , the moduli of the images of these points vary from 0² = 0 to (\sqrt{2})^{2} = 2. Thus, the image of this side is a vertical line segment from 0 to 2i contained in the v-axis and shown in black in Figure 2.22(b). In a similar manner, we find that the image of the side of S containing the vertices 0 and 1 − i is a vertical line segment from 0 to −2i contained in the v-axis. See Figure 2.22. The remaining side of S contains the vertices 1 − i and 1 + i. This side consists of the set of points z = 1+i y,  −1 ≤ y ≤ 1. Because this side is contained in the vertical line x = 1, it follows from (2) and (3) of Example 2 that its image is a parabolic segment given by:

u=k^{2}-y^{2},   v=2ky,  -\infty  \lt  y \lt  \infty      (2)

u=k^{2}- \frac{v^2}{4k^{2}},-\infty \lt v\lt \infty     (3)

Thus, we have shown that the image of triangle S shown in color in Figure 2.22(a) is the figure S^{\prime} shown in black in Figure 2.22(b).