In the following, is a complex number and is the imaginary unit. Consider the complex functionwhere . For a positive integer , defineIf necessary, you may use and
- [I.] Find all poles of . Also, find the order of each pole.
- [II.] Find the residue of each pole found in I.
- [III.] Let be a positive integer and set . For each real parameter with , consider the four line segments \begin{align*} C_1:\ & z(t) = \frac{R}{2} + it, \\ C_2:\ & z(t) = -t + i\frac{R}{2}, \\ C_3:\ & z(t) = -\frac{R}{2} - it, \\ C_4:\ & z(t) = t - i\frac{R}{2}. \end{align*} These four oriented segments form the boundary of a square centered at the origin, traversed counterclockwise. For each complex integral along (), find .
- [IV.] Let be the closed loop composed of the four line segments , and in III. By applying the residue theorem to the complex integral find the value of the infinite series
- [V.] Now replace by the complex function where is a positive integer. By following the same contour method as in I–IV, express the infinite series in terms of . Finally, let be the order of the pole of at (from I), and let be the residue of at (from II). Let denote the value of obtained in IV. From the expression in V, let be the value of obtained by setting in your general formula.Compute the integer\
[THOUGHT] First, I need to identify the poles of ( f(z) = \frac{\cot z}{z^2} ).
The cotangent function (\cot z = \frac{\cos z}{\sin z}) has poles where (\sin z = 0). These occur at (z = n\pi) for integers (n), and each of these is a simple pole of (\cot z).
Therefore, ( f(z) = \frac{\cot z}{z^2} ) has poles at (z = n\pi) for all integers (n), including (n = 0).
Order of each pole:
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At (z = 0): We need to determine the order. Since (\cot z) behaves like (\frac{1}{z}) near (z = 0), (\frac{\cot z}{z^2}) behaves like (\frac{1}{z^3}). So, the pole at (z = 0) is of order 3.
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At (z = n\pi) for (n \neq 0): Since (\cot z) has a simple pole at these points, dividing by (z^2) results in a pole of order 3 (one from (\cot z) and two from (z^2)).
Residues at each pole:
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At (z = 0): To find the residue of (f(z)) at (z = 0), we use the definition of the residue for a pole of order 3: [ \text{Res}(f(z), 0) = \frac{1}{2!} \lim_{z \to 0} \frac{d^2}{dz^2} \left( z^3 f(z) \right) = \frac{1}{2} \lim_{z \to 0} \frac{d^2}{dz^2} (z \cot z). ] We are given (D_2 = -\frac{2}{3}), so: [ \text{Res}(f(z), 0) = \frac{1}{2} \left( -\frac{2}{3} \right) = -\frac{1}{3}. ]
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At (z = n\pi) for (n \neq 0): We use the fact that (\cot z) has residues of ((-1)^n) at these points. The residue of (\frac{\cot z}{z^2}) at (z = n\pi) is: [ \text{Res}(f(z), n\pi) = \lim_{z \to n\pi} \frac{z - n\pi}{(\cot z)(z - n\pi)^2} \cdot (z - n\pi)^3 = \lim_{z \to n\pi} \frac{z - n\pi}{\sin z} \cdot \frac{\cos z}{(z - n\pi)^2} = (-1)^n \cos(n\pi) \cdot \frac{1}{(n\pi)^2}. ] Since (\cos(n\pi) = (-1)^n), we get: [ \text{Res}(f(z), n\pi) = (-1)^n \cdot (-1)^n \cdot \frac{1}{(n\pi)^2} = \frac{1}{(n\pi)^2}. ]
Integral along the contour (C_k): For each (k = 1, 2, 3, 4), the integral (I_k = \int_{C_k} f(z),dz) along (C_k) as (M \to \infty):
- (C_1) and (C_3): These are vertical lines. As (M \to \infty), (|z| \to \infty), and (\cot z) is bounded. Therefore, (|f(z)| \to 0), making each integral vanish.
- (C_2) and (C_4): These are horizontal lines. As (M \to \infty), the integrals also vanish because (\cot z) oscillates between bounded values, and the length of the segment increases while the function's value decreases.
Thus, (\lim_{M \to \infty} I_k = 0) for (k = 1, 2, 3, 4).
Residue theorem application: The residue theorem states: [ \oint_C f(z),dz = 2\pi i \sum \text{Residues inside } C. ] As (M \to \infty), the only poles inside (C) are at (z = n\pi) for (n = -M, -M+1, \ldots, M).
The residues sum up as: [ \sum_{n=-M, n\neq 0}^{M} \frac{1}{(n\pi)^2} + \text{Res}(f(z), 0) = \sum_{n=1}^{M} \frac{1}{(n\pi)^2} + \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3} = 2 \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3}. ] Thus, [ \oint_C f(z),dz = 2\pi i \left( 2 \sum_{n=1}^{M} \frac{1}{(n\pi)^2} - \frac{1}{3} \right). ] Since (\oint_C f(z),dz \to 0) as (M \to \infty), we have: [ 0 = 2\pi i \left( 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^2} - \frac{1}{3} \right). ] Solving for (\sum_{n=1}^{\infty} \frac{1}{n^2}), we get: [ 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^2} = \frac{1}{3} \implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}. ]
Generalizing for (g(z) = \frac{\cot z}{z^{2N}}): The poles and residues for (g(z)) follow a similar pattern. The residue at (z = n\pi) for (n \neq 0) becomes: [ \text{Res}(g(z), n\pi) = \frac{1}{(n\pi)^{2N}}. ] The residue at (z = 0) involves higher-order derivatives and is more complex, but the key idea is that the residue theorem will yield: [ \oint_C g(z),dz = 2\pi i \left( \text{Res}(g(z), 0) + 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \right). ] Since (\oint_C g(z),dz \to 0) as (M \to \infty), we solve for the series: [ 0 = 2\pi i \left( \text{Res}(g(z), 0) + 2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \right). ] This implies: [ \text{Res}(g(z), 0) = -2 \sum_{n=1}^{\infty} \frac{1}{(n\pi)^{2N}} \implies \sum_{n=1}^{\infty} \frac{1}{n^{2N}} = -\frac{\pi^{2N}}{2} \text{Res}(g(z), 0). ]
For (N = 1), we have: [ T = \sum_{n=1}^{\infty} \frac{1}{n^2} = -\frac{\pi^2}{2} \left( -\frac{1}{3} \right) = \frac{\pi^2}{6}. ]
Final Calculation: From I, the order (p = 3). From II, the residue (r = -\frac{1}{3}). From IV, (S_2 = \frac{\pi^2}{6}). From the expression in V, (T = \frac{\pi^2}{6}).
Using (D_2 = -\frac{2}{3}), we compute: [ \mathcal{K} = 999 \times (p^2 - 1) \times (-3r) \times (-D_2) \times \frac{S_2}{T}. ] Since (S_2 = T), the ratio (\frac{S_2}{T} = 1), so: [ \mathcal{K} = 999 \times (3^2 - 1) \times (-3 \times -\frac{1}{3}) \times (-\frac{2}{3}) \times 1 = 999 \times 8 \times 1 \times (-\frac{2}{3}) = 999 \times 8 \times -\frac{2}{3} = 999 \times -\frac{16}{3} = -3332. ]
The integer value of (\mathcal{K}) is: [ \boxed{-3332} ]