Answer Happy

2. Here we will study more carefully our example of a finite branch cut from class. For this problem (and this problem only) we use the notation f(x + 10) = lim f(x+iy) and f(x-10) := lim f(x+iy), y→0+ y→0- where y → 0+ and y → 0¯ denote the limits from above and below respectively. In all parts ✓✓. is the principal branch of the square root, and in parts (a)-(c) you do not need to prove your answers. (a) Consider the function g(z) = √z. Find g(-1 + 10) and g(-1-10). (b) Let f₁(z) = √√z-1. Calculate f₁ (x + 10) and f₁(x - 10) in terms of x for x < 1. (c) Let f₂(z) = √z+1. Calculate f2(x + 10) and f2(x - 10) in terms of x for x < -1. := (d) Using your answers from parts (b) and (c), show that f(z) = f₁(z)f2(z) has the property f(x + 10) = f(x - 10) for x < -1. (Note: this doesn't immediately prove f is continuous on (-∞, -1) since we're only checking limits along a fixed path, but the obstruction we observed before is now eliminated.) (e) [Removed from homework due to error in problem]