Question 7.2.3: Image of a Circle. Find the image of the unit circle |z| = 2......
Image of a Circle
Find the image of the unit circle |z| = 2 under the linear fractional transformation T(z) = (z + 2)/ (z − 1). What is the image of the disk |z| ≤ 2 under T?
In this example the pole z = 1 does not lie on the circle |z| = 2, and so Theorem 7.3 indicates that the image of |z| = 2 is a circle C^{\prime}. To find an algebraic description of C^{\prime}, we first note that the circle |z| = 2 is symmetric with respect to the x-axis. That is, if z is on the circle |z| = 2, then so is \overline{z} . Furthermore, we observe that for all z,
T(\bar{z})={\frac{\bar{z}+2}{\bar{z}-1}}={\frac{\overline{{z+2}}}{\overline{{z-1}}}}=\overline{{{\left({\frac{z+2}{z-1}}\right)}}}=\overline{{{T(z)}}}.Hence, if z and \overline{z} are on the circle |z| = 2, then we must have that both w\,=\,T(z)\,\mathrm{~and~}\bar{w}\,=\,\overline{{{T(z)}}}\,=\,T(\bar{z}) are on the circle C^{\prime}. It follows that C^{\prime} is symmetric with respect to the u-axis. Since z = 2 and −2 are on the circle |z| = 2, the two points T(2) = 4 and T(−2) = 0 are on C^{\prime}. The symmetry of C^{\prime} implies that 0 and 4 are endpoints of a diameter, and so C^{\prime} is the circle |w − 2| = 2. Using z = 0 as a test point, we find that w = T(0) = −2, which is outside the circle |w − 2| = 2. Therefore, the image of the interior of the circle |z| = 2 is the exterior of the circle |w − 2| = 2. In summary, the disk |z| ≤ 2 shown in color in Figure 7.14(a) is mapped onto the region |w − 2| ≥ 2 shown in gray in Figure 7.14(b) by the linear fractional transformation T(z) = (z + 2)/ (z − 1).
