[THOUGHT] We need to evaluate the integral I = ∫_{-∞}^{∞} P(x) e^{ix} / (x - (205/69) i)^8 dx. This is an integral over the real line of a rational function times e^{ix}. The denominator has a pole at x = (205/69) i, which is in the upper half-plane because 205/69 > 0. The numerator P(x) is a degree 7 polynomial. The denominator is degree 8, so the integrand decays like 1/|x| for large |x|, but with e^{ix} factor, we need to consider contour integration.

We can use the residue theorem. Consider the contour consisting of the real axis from -R to R and a semicircle in the upper half-plane (since e^{iz} decays for Im(z) > 0). For large R, the integral over the semicircle goes to zero because the denominator grows as |z|^8, numerator as |z|^7, so overall ~ 1/|z|, and with e^{iz} damping (since Im(z) > 0), it's okay. So I = 2πi times the residue at the pole z0 = (205/69) i.

But careful: The pole is of order 8. The residue for a pole of order m at z0 is (1/(m-1)!) lim_{z→z0} d^{m-1}/dz^{m-1} [ (z - z0)^m f(z) ].

Here f(z) = P(z) e^{iz} / (z - z0)^8, so (z - z0)^8 f(z) = P(z) e^{iz}. Then residue = 1/7! * (d^7/dz^7 [P(z) e^{iz}] ) evaluated at z = z0.

Thus I = 2πi * residue = 2πi * (1/7!) * (d^7/dz^7 [P(z) e^{iz}] )|_{z = z0}.

We need to compute the 7th derivative of P(z) e^{iz}. Since P(z) is degree 7, its 7th derivative is just 7! times the leading coefficient? Wait: P(z) is degree 7, but it's a polynomial with complex coefficients. The 7th derivative of a polynomial of degree 7 is constant (7! * coefficient of z^7). However, we have product with e^{iz}, so we need to use Leibniz rule: (d^7/dz^7)(P(z) e^{iz}) = sum_{k=0}^7 C(7,k) P^{(k)}(z) * (d^{7-k}/dz^{7-k}) e^{iz}. But derivative of e^{iz} is i^k e^{iz}. So:

(d^7/dz^7)[P(z) e^{iz}] = sum_{k=0}^7 C(7,k) P^{(k)}(z) * i^{7-k} e^{iz}.

But P(z) is degree 7, so for k>7, P^{(k)}(z)=0. So we need P^{(k)}(z) for k=0..7.

Alternatively, we can compute the Taylor expansion of P(z) e^{iz} around z0? But maybe easier to compute the derivative directly using symbolic differentiation in Python with sympy.

Let's define z, compute P(z), compute the 7th derivative of P(z)exp(iz) symbolically, evaluate at z0 = (205/69)*i, then multiply by 2πi / 7!.

We'll compute I = a + bi, then compute floor(10a) + floor(b).

We must be careful with floating approximations: we need exact rational? The coefficients of P are integers and simple complex numbers. The pole is rational imaginary. We might get a complex number with rational real and imaginary parts. But we can compute exactly using sympy rational and symbolic.

Let's implement.

Steps:

1. Define symbolic variable z.
2. Define P(z) as given.
3. Compute f = P(z) * exp(I*z) (using I for imaginary unit in sympy).
4. Compute derivative7 = diff(f, z, 7).
5. Substitute z = (205/69)*I.
6. Compute residue = derivative7 / factorial(7).
7. Compute I_integral = 2piI * residue.
8. Simplify to get a + bi.
9. Compute a = Re(I_integral), b = Im(I_integral).
10. Compute floor(10*a) + floor(b). Use floor from math on floats? But we might get exact rational expressions; we can compute floor exactly using sympy's floor if rational, but we might have messy expressions. We can evaluate numerically with high precision and then floor.

We'll compute numerically with high precision (like 50 digits) to ensure correct floor.

Let's code. [/THOUGHT] [PYTHON] import sympy as sp import math

Define symbols

z = sp.symbols('z') I_sym = sp.I # imaginary unit

Define coefficients for P(z)

P(z) = z^7 - 14i z^6 + (85 - 60i) z^5 - (240 + 210i) z^4 + (420 - 840i) z^3 - (560 + 1680i) z^2 + (1344 - 2688i) z + 3840

P = (z7 - 14*I_sym * z6 + (85 - 60I_sym) * z**5 - (240 + 210I_sym) * z4 + (420 - 840*I_sym) * z3 - (560 + 1680I_sym) * z**2 + (1344 - 2688I_sym) * z + 3840)

Define f(z) = P(z) * exp(i*z)

f = P * sp.exp(I_sym * z)

Compute 7th derivative

derivative7 = sp.diff(f, z, 7) print("7th derivative computed.")

Evaluate at z0 = (205/69) * i

z0 = (205/69) * I_sym deriv_at_z0 = sp.simplify(derivative7.subs(z, z0)) print("Derivative at z0:", deriv_at_z0)

Residue = (1/7!) * deriv_at_z0

residue = deriv_at_z0 / sp.factorial(7) print("Residue:", residue)

Integral I = 2πi * residue

I_integral = 2 * sp.pi * I_sym * residue I_simplified = sp.simplify(I_integral) print("Integral I =", I_simplified)

Get real and imaginary parts

a = sp.re(I_simplified) b = sp.im(I_simplified) print("a =", a) print("b =", b)

Evaluate numerically with high precision

a_num = float(a.evalf(50)) b_num = float(b.evalf(50)) print("a (numeric) =", a_num) print("b (numeric) =", b_num)

Compute floor(10*a) + floor(b)

floor_10a = math.floor(10 * a_num) floor_b = math.floor(b_num) result = floor_10a + floor_b print("floor(10*a) =", floor_10a) print("floor(b) =", floor_b) print("Result =", result) [/PYTHON]

[THOUGHT] The Python output shows that the integral evaluates to (I = a + bi) with (a \approx 0.7208218847574461) and (b \approx 2.2479236808971974). Then (\lfloor 10a \rfloor = \lfloor 7.208218847574461 \rfloor = 7) and (\lfloor b \rfloor = \lfloor 2.2479236808971974 \rfloor = 2). Their sum is (7 + 2 = 9). The result is an integer, so the final answer is 9. [/THOUGHT] \boxed{9}