Question 2.4.8: Image of a Circular Sector under w = z½. Find the image of ......

Let S^{\prime} denote the image of S under w=z^{1/2}. Since |z| ≤ 3 for points in S and since z^{1/2} takes the square root of the modulus of a point, we must have that \left|w\right|\leq \sqrt{3}   for points w in S^{\prime}. In addition, since π/2 ≤ arg(z) ≤ 3π/4 for points in S  and since z^{1/2} halves the argument of a point, it follows that π/4 ≤ arg(w) ≤ 3π/8 for points w in S^{\prime}. Therefore, we have shown that the set S shown in color in Figure 2.30(a) is mapped onto the set S^{\prime} shown in gray in Figure 2.30(b) by w = z^{1/2}.