Advanced Calculus fi..

558 Advanced Calculus, Fifth EditionIt remains to show that the integral on the right iiTindeed f (zoJ'. 2ni. Now, sincef (20) = const,where we integrate always on the circle lz - zol = R. Hence on the same path,Now lz - zol = R on the path, and since f (z) is continuous at zo, 1 f (z) - f (zo)l < Efor R < 6, for each preassigned E > 0. Hence by Theorem 7,Thus the absolute value of the integral can be made as small as desired by choosingR sufficiently small. But the integral has the same value for all choices of R. Thisis possible only if the integral is zero for all R. Hence the left-hand side of (8.51) iszero, and (8.50) follows.The integral formula (8.50) is remarkable in that it expresses the values of thefunction f (z) at points zo inside the curve C in terms of the values along C alone.If C is taken as a circle z = zo + ~ e", then (8.50) reduces to the following:I??-