[THOUGHT] We need to parse the problem carefully. It defines f(z) = cot(z)/z^2. Then D_m = limit as z→0 of d^m/dz^m (z cot z). They give D_2 = -2/3. Also, limit of (z - nπ)/sin(z) = (-1)^n for integer n.

Part I: Find all poles of f(z) and their order. f(z) = cot(z)/z^2 = cos(z)/(z^2 sin(z)). Poles occur where denominator z^2 sin(z) = 0. sin(z)=0 at z = nπ for integer n. At z=0, we have z^2=0 and sin(z)=0, so we need to examine order.

At z=0: cot(z) ~ 1/z - z/3 - z^3/45 - ... (Laurent series). Then z cot(z) ~ 1 - z^2/3 - z^4/45 - ... So f(z) = cot(z)/z^2 = (z cot(z))/z^3 ~ (1 - z^2/3 - ...)/z^3 = 1/z^3 - 1/(3z) + ... So at z=0, f has a pole of order 3? Let's check: cot(z) = cos(z)/sin(z). At z=0, sin(z) ~ z, cos(z)~1, so cot(z) ~ 1/z. Then f(z) ~ (1/z)/z^2 = 1/z^3. So indeed pole of order 3 at z=0.

At z = nπ, n ≠ 0 integer. sin(z)=0, cos(z) = (-1)^n ≠ 0. So cot(z) has a simple pole (since sin(z) has simple zero, cos(z) nonzero). Then f(z) = cot(z)/z^2. Since z^2 is nonzero at z=nπ (n≠0), the pole order of f at z=nπ is same as cot(z) there, i.e., simple pole.

Thus poles: z=0 (order 3), z=nπ for n ∈ ℤ{0} (order 1).

Part II: Residue at each pole.

First, residue at z=0 for f(z). Since pole order 3, we need coefficient of 1/z in Laurent expansion.

We can compute Laurent series of cot(z) around 0: cot(z) = 1/z - z/3 - z^3/45 - ... (standard). Then f(z) = cot(z)/z^2 = 1/z^3 - 1/(3z) - z/45 - ... So residue at z=0 is coefficient of 1/z: -1/3.

Thus r = -1/3.

Now residue at z = nπ, n ≠ 0 integer. Since simple pole, residue = lim_{z→nπ} (z - nπ) f(z) = lim_{z→nπ} (z - nπ) cot(z)/z^2.

But cot(z) = cos(z)/sin(z). sin(z) has simple zero at z=nπ. Using given limit: lim_{z→nπ} (z - nπ)/sin(z) = (-1)^n. Then lim_{z→nπ} (z - nπ) cot(z) = lim (z - nπ) cos(z)/sin(z) = cos(nπ) * lim (z - nπ)/sin(z) = (-1)^n * (-1)^n = 1. Because cos(nπ) = (-1)^n, and lim (z - nπ)/sin(z) = (-1)^n, product = (-1)^n * (-1)^n = 1.

Thus residue at z=nπ is lim (z - nπ) f(z) = [lim (z - nπ) cot(z)] / (nπ)^2 = 1 / (nπ)^2.

So residue at z=nπ is 1/(n^2 π^2). For n positive and negative? Since n integer nonzero, and residue formula uses n^2, same for n and -n.

Part III: Square contour with side length R = π(2M+1). For each C_k, we need limit as M→∞ of I_k = ∫_C_k f(z) dz.

We need to analyze f(z) = cot(z)/z^2. For large |z|, cot(z) is bounded? Actually cot(z) = i (e^{iz}+e^{-iz})/(e^{iz}-e^{-iz}) = i (e^{2iz}+1)/(e^{2iz}-1). For z = x+iy, if |y| large, cot(z) tends to ±i depending on sign of y. But our contour is square with vertices at (±R/2, ±R/2). As M→∞, R→∞, the square expands. However, f(z) decays like 1/z^2 for large |z| where cot(z) is bounded? But cot(z) may not be bounded everywhere; it has poles at nπ, but the square's horizontal sides are y = ±R/2, vertical sides x = ±R/2. As M large, these sides are at large |x| or |y|. We need to evaluate limit of integrals along each side.

We can maybe use the fact that f(z) = cot(z)/z^2, and cot(z) is periodic with period π. Possibly the integral over each side tends to zero as M→∞? Let's think.

But the problem likely expects that each I_k tends to 0 as M→∞. However, we need to compute limit for each k separately. But maybe they are not all zero? Let's check.

We can parameterize each C_k:

C1: z(t) = R/2 + i t, t from -R/2 to R/2? Wait: "z(t) = R/2 + i t" but t is real parameter? They say "for each real parameter t with -R/2 ≤ t ≤ R/2". So t runs from -R/2 to R/2. So C1: right vertical segment? Actually z = R/2 + i t, t from -R/2 to R/2. That's a vertical segment at x = R/2, y from -R/2 to R/2. But they said "oriented segments form boundary traversed counterclockwise". Let's check orientation: C1: z(t) = R/2 + i t, with t increasing? If t goes from -R/2 to R/2, then starting at t = -R/2: z = R/2 - i R/2 (bottom right), then as t increases to R/2, we go to z = R/2 + i R/2 (top right). That's upward vertical segment on right side. That's consistent with counterclockwise orientation if we go from bottom right to top right? Actually typical counterclockwise square: start at (-R/2, -R/2), go to (R/2, -R/2) (rightwards), then up to (R/2, R/2) (upwards), then left to (-R/2, R/2) (leftwards), then down to (-R/2, -R/2) (downwards). But here C1 is vertical upward at right side, so maybe they started at bottom right? Let's see definitions:

C1: z(t) = R/2 + i t, t from -R/2 to R/2. That's indeed vertical upward at right side.

C2: z(t) = -t + i R/2, t from -R/2 to R/2. Write z = -t + i R/2. As t increases from -R/2 to R/2, -t goes from R/2 to -R/2. So starting at t=-R/2: z = R/2 + i R/2 (top right), then as t increases, z moves leftwards along top edge to z = -R/2 + i R/2 (top left). So C2 is top horizontal leftwards.

C3: z(t) = -R/2 - i t, t from -R/2 to R/2. As t increases, -t decreases? Actually t from -R/2 to R/2: at t=-R/2, z = -R/2 + i R/2? Wait: z = -R/2 - i t. For t = -R/2, z = -R/2 - i(-R/2) = -R/2 + i R/2 (top left). For t = R/2, z = -R/2 - i(R/2) = -R/2 - i R/2 (bottom left). So C3 is vertical downward at left side.

C4: z(t) = t - i R/2, t from -R/2 to R/2. At t=-R/2: z = -R/2 - i R/2 (bottom left). At t=R/2: z = R/2 - i R/2 (bottom right). So C4 is bottom horizontal rightwards.

Thus orientation is indeed counterclockwise: right side up, top side left, left side down, bottom side right.

Now we need limit as M→∞ of I_k = ∫_C_k f(z) dz.

We can compute each integral explicitly? Possibly using symmetry? f(z) is even? f(-z) = cot(-z)/((-z)^2) = (-cot(z))/z^2 = -cot(z)/z^2 = -f(z). So f is odd. That might help.

But as M→∞, the sides go to infinity. For large |z|, f(z) ~ cot(z)/z^2. On vertical sides, x = ±R/2, y varies from -R/2 to R/2. As R→∞, |x| large, but cot(z) for large |x|? Actually cot(z) = i (e^{iz}+e^{-iz})/(e^{iz}-e^{-iz}) = i (e^{2iz}+1)/(e^{2iz}-1). For z = x+iy, e^{2iz} = e^{2i(x+iy)} = e^{2ix - 2y}. So magnitude |e^{2iz}| = e^{-2y}. So if y is fixed and |x|→∞, e^{2iz} oscillates, but its magnitude depends only on y. So on vertical sides, x is large but fixed ±R/2, but R→∞, so x→∞, but magnitude still depends on y. However, as |x|→∞, e^{2iz} oscillates rapidly, but the ratio (e^{2iz}+1)/(e^{2iz}-1) is bounded? Actually for large |x|, e^{2iz} is oscillatory, but its magnitude is e^{-2y}. So if y>0, e^{-2y} <1, so |e^{2iz}| small, then (e^{2iz}+1)/(e^{2iz}-1) ≈ -1, so cot(z) ≈ -i? Wait, careful.

Better approach: Use known integral of cot(z)/z^2 over large contours. Possibly the integral over the whole square tends to 0 as M→∞? But we need limit of each side individually.

Maybe the problem expects that each I_k tends to 0 as M→∞. However, later part IV uses residue theorem to find sum 1/n^2. That suggests the integral over the whole square tends to something nonzero? Actually by residue theorem, ∮C f(z) dz = 2πi * sum of residues inside C. As M→∞, the square encloses more poles at nπ, ±nπ, etc. The sum of residues inside is sum{n=-M}^{M} 1/(n^2 π^2) (excluding n=0? Actually n=0 pole is inside, residue -1/3). Wait, we have poles at z=nπ, n integer. Inside the square, as M→∞, we have poles at nπ for n = -M,...,M? But careful: square side R = π(2M+1). The square's right side at x = R/2 = π(2M+1)/2. So the poles on real axis are at nπ. The largest positive pole inside is less than R/2? Since R/2 = π(2M+1)/2 = πM + π/2. So poles at nπ with n up to M? Because nπ ≤ πM + π/2 implies n ≤ M + 0.5, so n ≤ M. Similarly negative poles down to -M. So inside we have poles at z = nπ for n = -M, -M+1, ..., M. That's 2M+1 poles plus z=0. But z=0 is also a pole (order 3). So total poles inside: 2M+1 + 1 = 2M+2 poles? Wait, n from -M to M inclusive gives 2M+1 poles, plus z=0 which is n=0 already counted? Actually n=0 is included in the list from -M to M. So total distinct poles: 2M+1 poles (including n=0). But n=0 is triple pole, others simple.

Thus sum of residues inside = residue at z=0 + sum_{n=-M, n≠0}^{M} 1/(n^2 π^2) = -1/3 + (2/π^2) * sum_{n=1}^{M} 1/n^2.

Now as M→∞, ∮C f(z) dz = 2πi * (-1/3 + (2/π^2) S_M) where S_M = sum{n=1}^{M} 1/n^2.

But also ∮_C f(z) dz = sum of integrals over four sides: I1+I2+I3+I4.

As M→∞, each I_k might have a limit. Possibly each I_k tends to 0? Then total integral tends to 0, implying -1/3 + (2/π^2) S∞ = 0? That would give S∞ = π^2/6, which is correct. So indeed total integral tends to 0 as M→∞. But we need to verify that each I_k individually tends to 0.

Let's compute limit of I_k. Since f(z) decays like 1/|z|^2 for large |z| (because cot(z) is bounded on the contour? Actually on vertical sides, x large, but cot(z) may not be bounded uniformly? Let's examine magnitude.

For large |z|, if Im(z) > 0, cot(z) → -i; if Im(z) < 0, cot(z) → i. But our vertical sides are at x = ±R/2, y from -R/2 to R/2. So on the top half of vertical side, y positive, cot(z) ~ -i; on bottom half, y negative, cot(z) ~ i. However, as R→∞, y ranges over finite interval, so cot(z) is bounded. Indeed, for fixed y, as |x|→∞, cot(z) tends to ±i depending on sign of y. But more precisely, |cot(z)| is bounded by some constant independent of x? Actually for large |x|, |cot(z)| is bounded by something like 2/(1-e^{-2|y|})? Let's compute: cot(z) = i (e^{2iz}+1)/(e^{2iz}-1). Let w = e^{2iz} = e^{2ix - 2y}. Then |w| = e^{-2y}. For y>0, |w|<1. Then (w+1)/(w-1) has magnitude bounded by (|w|+1)/(1-|w|). So |cot(z)| = |(w+1)/(w-1)| ≤ (e^{-2y}+1)/(1-e^{-2y}) = (1+e^{-2y})/(1-e^{-2y}) = coth(y). Similarly for y<0, |w|>1, then |cot(z)| = |(w+1)/(w-1)| = |(1+1/w)/(1-1/w)| ≤ (1+e^{2y})/(1-e^{2y})? Actually symmetry: cot(z) is odd in y? Not exactly. But we can bound |cot(z)| by some constant depending on y. Since y ∈ [-R/2, R/2], as R→∞, y can be arbitrarily large in magnitude, but on vertical sides, y is bounded by R/2, which goes to infinity. However, for large |y|, cot(z) tends to ±i, so bounded by 2 maybe. So overall, on the entire vertical side, |cot(z)| is bounded by some constant C (maybe depending on R? but uniformly for all R? Actually for large |y|, bound tends to 1). So we can say there exists a constant C such that |cot(z)| ≤ C for all z on those sides, independent of R? Not exactly, because as |y| increases, bound may increase? Let's compute bound: For y>0, |cot(z)| ≤ coth(y). coth(y) decreases to 1 as y→∞. So maximum over y≥0 is at y=0, but at y=0, cot(z) has poles, but our vertical side does not include y=0? Actually y=0 is included (since y from -R/2 to R/2). At y=0, cot(z) = cot(x) which is unbounded as x→∞? Wait, at y=0, cot(z) = cot(x) where x = ±R/2, which is large, but cot(x) oscillates and is unbounded? Actually cot(x) has poles at integer multiples of π, but for large x not near poles, cot(x) is bounded? No, cot(x) is not bounded as x→∞ because it has poles spaced by π, so near those poles it becomes arbitrarily large. But our vertical side is at x = R/2, which as R→∞ goes to infinity, but R/2 is not necessarily near a pole? However, we can choose R = π(2M+1). Then R/2 = π(2M+1)/2 = πM + π/2. That is not an integer multiple of π; it's π/2 mod π. So x = πM + π/2. Then cot(x) = cot(πM + π/2) = cot(π/2) = 0. So at y=0, cot(z) = cot(πM + π/2) = 0. So no pole on the vertical line. Similarly, on horizontal lines y = ±R/2, we have y = ±π(2M+1)/2, which is also π/2 mod π. So sin(y) = ±1? Actually sin(y) = sin(π(2M+1)/2) = sin(πM + π/2) = (-1)^M? Not important. But cot(z) on horizontal line: z = x ± i R/2, so cot(z) = cos(z)/sin(z). For large |y|, cot(z) tends to ±i, so bounded.

Thus on the entire contour, |cot(z)| is bounded by some constant independent of R? Possibly yes, because the contour avoids poles. We can find a uniform bound: For any z on the square, |cot(z)| ≤ C for some C? Let's test numerically for large M. But we can argue: On the vertical sides, x = ±R/2 = ±π(2M+1)/2. Then z = ±π(2M+1)/2 + iy. Write cot(z) = i (e^{2iz}+1)/(e^{2iz}-1). Compute e^{2iz} = e^{2i(±π(2M+1)/2 + iy)} = e^{±iπ(2M+1) - 2y} = e^{±iπ(2M+1)} e^{-2y}. Since e^{±iπ(2M+1)} = e^{iπ(2M+1)} for plus sign, and = e^{-iπ(2M+1)} for minus sign. But e^{iπ(2M+1)} = e^{iπ} e^{i2πM} = -1 * 1 = -1. Similarly e^{-iπ(2M+1)} = -1. So e^{2iz} = - e^{-2y} for right side (x = R/2). For left side (x = -R/2), e^{2iz} = - e^{-2y} as well? Let's compute: x = -R/2 = -π(2M+1)/2. Then e^{2iz} = e^{2i(-π(2M+1)/2 + iy)} = e^{-iπ(2M+1) - 2y} = e^{-iπ(2M+1)} e^{-2y} = (-1) e^{-2y}. So same: e^{2iz} = - e^{-2y}. So interestingly, on both vertical sides, e^{2iz} = - e^{-2y}. Then cot(z) = i ( - e^{-2y} + 1 )/( - e^{-2y} - 1 ) = i (1 - e^{-2y}) / (-(1+ e^{-2y})) = i (1 - e^{-2y}) / -(1+ e^{-2y}) = -i (1 - e^{-2y})/(1+ e^{-2y}) = -i tanh(y). Because tanh(y) = (e^y - e^{-y})/(e^y + e^{-y}) = (1 - e^{-2y})/(1+ e^{-2y}). So cot(z) = -i tanh(y) on right vertical side. Similarly, on left vertical side? We got same e^{2iz} = - e^{-2y}. So cot(z) = -i tanh(y) as well? Let's verify: For left vertical side, x = -R/2, we have e^{2iz} = - e^{-2y}. Then cot(z) = i (e^{2iz}+1)/(e^{2iz}-1) = i ( - e^{-2y} + 1 )/( - e^{-2y} - 1 ) = i (1 - e^{-2y})/( - (1+ e^{-2y})) = -i (1 - e^{-2y})/(1+ e^{-2y}) = -i tanh(y). So indeed on both vertical sides, cot(z) = -i tanh(y). That's neat!

Similarly, on horizontal sides: y = ±R/2 = ±π(2M+1)/2. Let's compute e^{2iz} = e^{2i(x ± i R/2)} = e^{2ix ∓ R}. Since R = π(2M+1), so e^{2iz} = e^{2ix} e^{∓ R}. For top side y = R/2, e^{2iz} = e^{2ix} e^{-R}. For bottom side y = -R/2, e^{2iz} = e^{2ix} e^{R}. Then cot(z) = i (e^{2iz}+1)/(e^{2iz}-1). For top side, since e^{2iz} is very small (magnitude e^{-R} extremely small), we have cot(z) ≈ i (1)/( -1 ) = -i. More precisely, cot(z) = i (e^{2ix} e^{-R} + 1)/(e^{2ix} e^{-R} - 1). As R→∞, e^{-R}→0, so cot(z) → i (0+1)/(0-1) = -i. Similarly, on bottom side, e^{2iz} is huge, so cot(z) ≈ i (e^{2ix} e^{R})/(e^{2ix} e^{R}) = i. Actually more precisely, for large e^{R}, (e^{2iz}+1)/(e^{2iz}-1) ≈ 1, so cot(z) ≈ i. Wait check: For bottom side, e^{2iz} = e^{2ix} e^{R}, magnitude large. Then (e^{2iz}+1)/(e^{2iz}-1) = (1 + e^{-2iz})/(1 - e^{-2iz}) ≈ 1, so cot(z) ≈ i. So on bottom side, cot(z) ≈ i.

But we need exact expression? Possibly we can bound |cot(z)| on horizontal sides by some constant.

Now compute integrals:

I1 = ∫{C1} f(z) dz = ∫{t=-R/2}^{R/2} f(R/2 + i t) i dt.

f(z) = cot(z)/z^2. On C1, cot(z) = -i tanh(t). So f(z) = -i tanh(t) / ( (R/2 + i t)^2 ). Then I1 = i ∫ dt * (-i tanh(t)) / ( (R/2 + i t)^2 ) = ∫ dt tanh(t) / ( (R/2 + i t)^2 ). Because i * (-i) = 1.

Thus I1 = ∫_{-R/2}^{R/2} tanh(t) / ( (R/2 + i t)^2 ) dt.

As R→∞, denominator ~ (R/2)^2 for large R? Actually (R/2 + i t)^2 = (R/2)^2 + i R t - t^2. For large R, this is ~ (R/2)^2. So integrand ~ tanh(t) / (R^2/4) = (4 tanh(t))/R^2. Since tanh(t) is odd, the integral over symmetric interval might be small? Actually tanh(t) is odd, but the integrand is tanh(t) / ( (R/2)^2 + ... ). Since denominator is even in t? Let's check: (R/2 + i t)^2 is not even in t; its real part is (R/2)^2 - t^2, imaginary part is R t. So the integrand is complex. But magnitude ~ O(1/R^2). So integral over t from -R/2 to R/2: length ~ R, so magnitude ~ O(1/R). So I1 → 0 as R→∞.

Similarly, I3 (left vertical side) also tends to 0.

Now I2: C2: z(t) = -t + i R/2, t from -R/2 to R/2. Then dz = -dt (since derivative w.r.t t is -1). Actually z(t) = -t + i R/2, so dz/dt = -1, so dz = -dt. Then I2 = ∫{t=-R/2}^{R/2} f(-t + i R/2) (-dt) = -∫{-R/2}^{R/2} f(-t + i R/2) dt.

But we can use symmetry: f is odd, so f(-z) = -f(z). Let w = t - i R/2? Not sure.

Alternatively, compute directly using cot expression on horizontal side. On top side (y = R/2), we have cot(z) ≈ -i for large R, but more precisely, we have exact expression? Earlier we derived e^{2iz} = e^{2ix} e^{-R}. Since R large, e^{-R} tiny. Then cot(z) = i (e^{2ix} e^{-R} + 1)/(e^{2ix} e^{-R} - 1). For large R, this is approximately -i, but we can bound its magnitude: |cot(z)| ≤ something.

But maybe easier: Since f(z) decays like 1/|z|^2, and the length of each side is O(R), the integral magnitude is O(1/R) → 0. So each I_k → 0 as M→∞.

Thus limit of each I_k is 0.

But wait, part III asks: "For each complex integral I_k = ∫C_k f(z) dz along C_k (k=1,2,3,4), find lim{M→∞} I_k." Likely answer is 0 for each.

Thus we can state lim I_k = 0.

Now part IV: Using residue theorem on ∮C f(z) dz = sum of residues inside. As M→∞, sum of residues inside = -1/3 + (2/π^2) S_M, where S_M = sum{n=1}^M 1/n^2. And ∮_C f(z) dz = I1+I2+I3+I4 → 0+0+0+0 = 0. Thus -1/3 + (2/π^2) S∞ = 0, so S∞ = π^2/6.

Thus S_2 = π^2/6.

Part V: g(z) = cot(z)/z^{2N}. They want sum_{n=1}^∞ 1/n^{2N} expressed in terms of D_m. They give D_m = lim_{z→0} d^m/dz^m (z cot z). For N=1, we have f(z) = cot(z)/z^2. Then sum 1/n^2 = π^2/6. In terms of D_m, we need to express sum 1/n^{2N} using D_m.

We need to derive general formula. For g(z) = cot(z)/z^{2N}, the poles: at z=0 pole of order 2N+1? Let's check: cot(z) ~ 1/z - z/3 - ... So g(z) ~ 1/z^{2N+1} - 1/(3 z^{2N-1}) - ... So pole order = 2N+1 at z=0. At z=nπ, simple pole as before.

Residue at z=nπ is 1/(n^{2N} π^{2N})? Wait: residue at z=nπ for g(z) = lim_{z→nπ} (z - nπ) g(z) = lim (z - nπ) cot(z)/z^{2N} = [lim (z - nπ) cot(z)] / (nπ)^{2N} = 1/(n^{2N} π^{2N}) as before.

Now consider contour integral ∮C g(z) dz. As M→∞, sum of residues inside = residue at z=0 + sum{n=-M, n≠0}^{M} 1/(n^{2N} π^{2N}) = Res(g,0) + (2/π^{2N}) S_M^{(2N)} where S_M^{(2N)} = sum_{n=1}^M 1/n^{2N}.

Now we need Res(g,0). Since g(z) = cot(z)/z^{2N} = (z cot(z))/z^{2N+1}. Write h(z) = z cot(z). Then g(z) = h(z)/z^{2N+1}. We know Laurent expansion of h(z) around 0: h(z) = 1 - (1/3) z^2 + ... (since D_2 = limit_{z→0} h''(z) = -2/3? Actually D_m = limit_{z→0} d^m/dz^m (z cot z). So D_m = h^{(m)}(0). Because h is analytic at 0? Let's check: h(z) = z cot(z) is analytic at 0 (since z cot(z) → 1). So h(z) has a Taylor series: h(z) = sum_{k=0}^∞ a_k z^k, with a_0 = 1, a_1 = 0, a_2 = D_2/2! ? Wait: D_m = h^{(m)}(0). So a_m = D_m / m!.

Thus h(z) = 1 + (D_1/1!) z + (D_2/2!) z^2 + (D_3/3!) z^3 + ... But D_1 = limit_{z→0} h'(z). Compute h'(z) = cot(z) - z csc^2(z). At z=0, limit? Use series: cot(z) = 1/z - z/3 - ..., so h'(z) = (1/z - z/3 - ...) - z (1/z^2 + ...)? Actually csc^2(z) = 1/z^2 + 1/3 + ... So z csc^2(z) = 1/z + ... So h'(z) = (1/z - z/3 - ...) - (1/z + ...) = -z/3 + ... So h'(0) = 0. So D_1 = 0. Similarly D_2 = -2/3 as given.

Thus h(z) = 1 + 0*z + (D_2/2!) z^2 + ... = 1 - (1/3) z^2 + ... (since D_2 = -2/3, D_2/2! = (-2/3)/2 = -1/3). Good.

Now g(z) = h(z)/z^{2N+1}. So residue at z=0 for g is coefficient of 1/z in expansion of g(z). That is coefficient of z^{2N} in h(z). Because g(z) = h(z) * z^{-(2N+1)}. Write h(z) = sum_{m=0}^∞ a_m z^m, a_m = D_m/m!. Then g(z) = sum_{m=0}^∞ a_m z^{m - 2N -1}. The term with exponent -1 occurs when m - 2N -1 = -1 => m = 2N. So residue at z=0 is a_{2N} = D_{2N}/(2N)!.

Thus Res(g,0) = D_{2N}/(2N)!.

Now by residue theorem, ∮C g(z) dz = 2πi [ Res(g,0) + sum{n=-M, n≠0}^{M} Res(g, nπ) ] = 2πi [ D_{2N}/(2N)! + (2/π^{2N}) S_M^{(2N)} ].

As M→∞, the integral over the square tends to 0 (similar argument as before, because g(z) decays like 1/|z|^{2N} and |cot(z)| bounded). So ∮_C g(z) dz → 0. Thus we have:

D_{2N}/(2N)! + (2/π^{2N}) S∞^{(2N)} = 0, where S∞^{(2N)} = sum_{n=1}^∞ 1/n^{2N}.

Thus sum_{n=1}^∞ 1/n^{2N} = - (π^{2N} D_{2N})/(2 (2N)!).

But wait sign: D_{2N}/(2N)! + (2/π^{2N}) S = 0 => S = - (π^{2N} D_{2N})/(2 (2N)!).

For N=1, S_2 = sum 1/n^2 = - (π^2 D_2)/(2 * 2!) = - (π^2 * (-2/3))/(4) = (π^2 * 2/3)/4 = (π^2 * 2)/(12) = π^2/6. Indeed matches.

Thus T = S_2 = π^2/6.

Now we need p = order of pole of f(z) at z=0 from I. f(z) = cot(z)/z^2, we determined order = 3. So p = 3.

r = residue of f(z) at z=0 from II = -1/3.

D_2 = -2/3 (given).

S_2 = π^2/6.

T = sum_{n=1}^∞ 1/n^2 from setting N=1 in general formula = π^2/6.

Thus T = S_2.

Now compute 𝒦 = 999 * (p^2 - 1) * (-3r) * (-D_2) * (S_2 / T).

First compute p^2 - 1 = 9 - 1 = 8.

-3r = -3 * (-1/3) = 1.

-D_2 = -(-2/3) = 2/3.

S_2 / T = (π^2/6) / (π^2/6) = 1.

Thus product = 999 * 8 * 1 * (2/3) * 1 = 999 * 8 * (2/3) = 999 * (16/3) = (999/3)*16 = 333 * 16 = 5328.

Check: 333*16 = 5328.

Thus 𝒦 = 5328.

We should verify with Python to ensure no mistakes.

Let's compute using Python:

p = 3 r = -1/3 D2 = -2/3 S2 = pi^2/6 T = pi^2/6 Compute expression.

We'll use sympy for exact rational.

But note: S2/T = 1 exactly, so we can compute rationally.

Compute step by step.

We'll write code. </think>

[PYTHON] import sympy as sp

Define values

p = 3 r = sp.Rational(-1, 3) # -1/3 D2 = sp.Rational(-2, 3) # -2/3 S2 = sp.pi2 / 6 T = sp.pi2 / 6

Compute expression

factor1 = p**2 - 1 # 9-1=8 factor2 = -3 * r # -3 * (-1/3) = 1 factor3 = -D2 # -(-2/3) = 2/3 factor4 = S2 / T # = 1

K = 999 * factor1 * factor2 * factor3 * factor4 print(K) print(K.evalf()) [/PYTHON]

[THOUGHT] The Python output shows K = 5328. Therefore, the final answer is 5328.

\boxed{5328}