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] To solve the problem, we analyze each part step-by-step:
I. Poles and Orders:
- The function ( f(z) = \frac{\cot z}{z^2} ) has poles where ( \cot z ) has poles (at ( z = n\pi ) for integers ( n )) and at ( z = 0 ) due to the ( z^2 ) term.
- For ( n \neq 0 ), ( \cot z ) has simple poles, and dividing by ( z^2 ) (non-zero at ( z = n\pi )) retains the order 1.
- At ( z = 0 ), ( \cot z ) has a simple pole, and dividing by ( z^2 ) results in a pole of order 3.
II. Residues:
- For ( n \neq 0 ), the residue at ( z = n\pi ) is ( \frac{1}{(n\pi)^2} ).
- At ( z = 0 ), using the Laurent series expansion, the residue is ( -\frac{1}{3} ).
III. Integrals ( I_k ):
- As ( M \to \infty ), the integrals over each segment ( C_k ) tend to 0 due to the decay of the integrand.
IV. Series Sum ( S_2 ):
- Applying the residue theorem to the closed contour integral, the sum ( \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} ).
V. General Series ( S_{2N} ):
- For ( g(z) = \frac{\cot z}{z^{2N}} ), the residue at ( z = 0 ) is ( \frac{D_{2N}}{(2N)!} ). Using the residue theorem, the series sum is: [ \sum_{n=1}^\infty \frac{1}{n^{2N}} = -\frac{D_{2N} \pi^{2N}}{2(2N)!} ]
- For ( N = 1 ), this matches ( S_2 = \frac{\pi^2}{6} ).
Final Calculation of ( \mathcal{K} ):
- Given ( p = 3 ), ( r = -\frac{1}{3} ), ( D_2 = -\frac{2}{3} ), ( S_2 = \frac{\pi^2}{6} ), and ( T = \frac{\pi^2}{6} ): [ \mathcal{K} = 999 \cdot (3^2 - 1) \cdot (-3 \cdot -\frac{1}{3}) \cdot (-\frac{2}{3}) \cdot \frac{\frac{\pi^2}{6}}{\frac{\pi^2}{6}} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} = 5328 ]
[PYTHON]
Calculation of K
p = 3 r = -1/3 D2 = -2/3 S2 = (3.141592653589793**2)/6 T = S2 # Since S2/T = 1
K = 999 * (p**2 - 1) * (-3 * r) * (-D2) * (S2 / T) print(int(K)) [/PYTHON]
[THOUGHT] The Python code output confirms the calculation of ( \mathcal{K} = 5328 ). This value is derived from the given parameters and the derived series sum, consistent with the problem's requirements. [/THOUGHT] \boxed{5328}