Answer Happy

d), e) please
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-i0) := 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 + i0) and g(-1-10). (b) Let fi(z) = √z-1. Calculate fi (x + i0) and f₁(x − i0) in terms of x for x < 1. 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) Prove that h(z) = z² - 1 has the property h (C\ [-1,1]) CC\(-∞,0]. Use this to finally prove that f= √z² - 1 is continuous on C\ [-1,1].