Question

Suppose that f(z) has a pole of order m at z=z_0, and is otherwise analytic near z_0. Show that f'(z) also has a pole at z_0. What is the order of the pole and what is the residue there?

Answer 1

f(z) has a pole of order m at z=z_0 implies that f'(z) also has a pole, but its order is m+1 at z=z_0. The residue of f'(z) at the pole can be found by isolating the coefficient of (z - z_0)^{-1} in the Laurent series expansion of f'(z) around z_0.

Step-by-step explanation:

The question asks about complex analysis and specifically about how the derivative of a complex function behaves at a point where the original function has a pole. Given that f(z) has a pole of order m at z=z_0, it means that in the vicinity of z_0, f(z) can be expressed as:

f(z) = g(z)/(z - z_0)^m

Here, g(z) is analytic and nonzero at z_0.

The derivative of f(z), denoted as f'(z), near this pole is:

f'(z) = (g'(z)(z - z_0)^m - m*g(z)(z - z_0)^(m-1))/((z - z_0)^{2m})

When simplified, this demonstrates that f'(z) has a pole of order m+1 because the highest power in the denominator after simplification will be (z - z_0)^(2m) and the highest power in the numerator that includes z - z_0 is at most (m-1).

The residue of f'(z) at the pole z=z_0 is the coefficient of (z - z_0)^{-1} in the Laurent series expansion of f'(z) around z_0. This requires calculating the Laurent series and isolating the given term, which can be complex depending on the function g(z).