answer:The function f(z) = 1 / (1 - z)^2 has a pole of order 2 at z = 1. To find the residue, we can use the formula for the residue of a pole of order n, which is (-1)^(n-1) / (n-1)! times the (n-1)th derivative of the numerator divided by the nth derivative of the denominator, evaluated at the pole. However, in this case, we can use a simpler approach. Since the pole is of order 2, we can expand the function in a Laurent series around z = 1. We can rewrite f(z) as 1 / (1 - z)^2 = 1 / (-1)^2 (z - 1)^2 = 1 / (z - 1)^2. Now, the Laurent series for 1 / (z - 1)^2 is simply the series 1 / (z - 1)^2, which has a single term. The residue is the coefficient of the 1 / (z - 1) term, which is zero. However, this is not the entire story. A more careful analysis reveals that the function has a double pole at z = 1, and the residue of a double pole is defined as the coefficient of the 1 / (z - 1) term in the Laurent series expansion around the pole. However, when the function is expanded in a Laurent series around z = 1, the leading term is 1 / (z - 1)^2, and there is no 1 / (z - 1) term. However, we also note that as z approaches 1, (z - 1) approaches zero, and using L'Hôpital's rule or a direct series expansion, we see that (z - 1) * 1 / (z - 1)^2 approaches infinity, but (z - 1)^2 * 1 / (z - 1)^2 approaches 1. However we are trying to find the coefficient of the term 1 / (z-1). The coefficient of the term 1/(z-1) is actually undefined in this function as it is a double pole.

answer:Since the function f(z) = 1 / (1 - z)^2 has a pole of order 2 at z = 1, we can conclude that the residue of this function at z = 1 is actually undefined, as the residue is only defined for simple poles (poles of order 1) or as the coefficient of the 1 / (z - a) term in the Laurent series expansion around the pole a. However, in general, when a function has a pole of order n, the residue can be thought of as the coefficient of the (n-1)th term of the expansion. But in the traditional sense, the residue is not defined for poles of order greater than 1. But if we must assign a value to it, it's infinity or we could say it doesn't exist (more precise), but there are generalized residue that could be defined, like when integrating a function over a closed curve around a pole of order n > 1.

answer:<tool_call>[{name: fetch_manga, arguments: {page: 4, genres: Romance, School}}, {name: downloadminecraftjson, arguments: {minecraftversion: 1.15.2}}]</tool_call>

answer:To convert a character to its corresponding binary representation, we can utilize the ASCII character encoding table. This table maps each character to a unique value between 0 and 255, which can be represented as an 8-bit binary string. Here is the implementation of the `char_to_binary` function: ```python def char_to_binary(char: str) -> str: assert len(char) == 1, Input must be a single character ascii_val = ord(char) binary_str = bin(ascii_val)[2:].zfill(8) return binary_str ```