[THOUGHT] We need to evaluate the integral: [ I = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x - 3i)^8} dx ] where (P(x)) is a degree-7 polynomial with complex coefficients.

We can use contour integration in the complex plane. The integrand is (f(z) = \frac{P(z) e^{iz}}{(z - 3i)^8}). The pole is at (z = 3i), which is in the upper half-plane. Since we have (e^{iz}), for (z = x + iy), (e^{iz} = e^{i(x+iy)} = e^{ix} e^{-y}). For (y > 0), (e^{-y}) decays as (y \to +\infty). So we can close the contour in the upper half-plane (since (e^{iz}) decays in the upper half-plane? Actually, check: (e^{iz} = e^{i(x+iy)} = e^{ix} e^{-y}). For (y > 0), (e^{-y}) decays, yes. So we can close in the upper half-plane.

The contour is the real axis plus a semicircle at infinity in the upper half-plane. The integral over the semicircle vanishes as radius (R \to \infty) because (e^{iz}) decays exponentially and (|P(z)|/(|z-3i|^8) \sim |z|^7/|z|^8 = 1/|z|) for large (|z|), so the integrand decays like (1/R), and the length is (\pi R), so integral tends to 0? Actually, need to check: (|e^{iz}| = e^{-y}). On the semicircle, (z = R e^{i\theta}), (y = R \sin\theta), so (|e^{iz}| = e^{-R \sin\theta}). For (\theta \in [0,\pi]), (\sin\theta \ge 0), so the maximum is at (\theta = \pi/2), where (\sin\theta = 1), giving decay (e^{-R}). So the integrand decays exponentially in the upper half-plane. So the semicircle integral vanishes as (R \to \infty).

Thus, by the residue theorem, the integral equals (2\pi i) times the residue at (z = 3i).

So we need to compute: [ I = 2\pi i \cdot \operatorname{Res}_{z=3i} \frac{P(z) e^{iz}}{(z - 3i)^8}. ]

Since the pole is of order 8, the residue is given by: [ \operatorname{Res}{z=3i} f(z) = \frac{1}{7!} \lim{z \to 3i} \frac{d^7}{dz^7} \left[ (z - 3i)^8 f(z) \right]. ] But ((z - 3i)^8 f(z) = P(z) e^{iz}).

Thus: [ \operatorname{Res}{z=3i} f(z) = \frac{1}{7!} \left. \frac{d^7}{dz^7} \left( P(z) e^{iz} \right) \right|{z=3i}. ]

So we need to compute the 7th derivative of (P(z) e^{iz}) evaluated at (z = 3i).

We can compute this using Leibniz's rule for the 7th derivative of a product: [ \frac{d^7}{dz^7} [P(z) e^{iz}] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) \frac{d^{7-k}}{dz^{7-k}} e^{iz}. ] But (\frac{d^{m}}{dz^{m}} e^{iz} = i^m e^{iz}).

So: [ \frac{d^7}{dz^7} [P(z) e^{iz}] = e^{iz} \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) i^{7-k}. ] Evaluate at (z = 3i).

Thus: [ I = 2\pi i \cdot \frac{1}{7!} \left. e^{iz} \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) i^{7-k} \right|_{z=3i}. ]

We need to compute (P(z)) and its derivatives up to order 7 at (z = 3i). Since (P) is a degree-7 polynomial, (P^{(7)}(z)) is constant (the leading coefficient times (7!)). But we can compute everything numerically using Python.

Let's implement in Python.

We'll define the polynomial coefficients from highest degree to constant term. The polynomial is: [ P(x) = x^7 - 14i x^6 + (85 - 60i) x^5 - (240 + 210i) x^4 + (420 - 840i) x^3 - (560 + 1680i) x^2 + (1344 - 2688i) x + 3840. ]

We'll compute derivatives symbolically or numerically? Since we need derivatives at (z = 3i), we can compute directly by evaluating the polynomial and its derivatives using symbolic differentiation or using a finite difference? Better to use symbolic differentiation with sympy for exactness, but we only need a numeric result for (a) and (b). However, we must be careful with rounding. Since the final answer is integer floors, we need sufficient precision.

We can use sympy to compute derivatives exactly (with complex coefficients) and then evaluate at (3i), then compute the sum, multiply by factors, and get a complex number. Then compute (a) and (b) as real and imaginary parts.

But note: The integral result (I) is (a+bi). Then we compute (\lfloor 10a \rfloor + \lfloor b \rfloor). So we need (a) and (b) accurately enough to determine floors.

Let's code.

First, import sympy. Define the polynomial with complex coefficients. Compute derivatives up to 7. Then compute the sum. Then multiply by (2\pi i / 7!). Then get real and imaginary parts.

We'll print (a) and (b) with high precision.

Also, we must ensure that the semicircle contribution vanishes; we already argued yes.

Let's proceed. </think>

[PYTHON] import sympy as sp

Define variable

z = sp.symbols('z', complex=True)

Define polynomial P(z) with complex coefficients

Coefficients from highest degree to constant term

coeffs = [ 1, # x^7 -14sp.I, # x^6 (85 - 60sp.I), # x^5 -(240 + 210sp.I), # x^4 (420 - 840sp.I), # x^3 -(560 + 1680sp.I), # x^2 (1344 - 2688sp.I), # x^1 3840 # constant ]

Build polynomial

P = sum(c * z**(7 - i) for i, c in enumerate(coeffs))

Compute derivatives up to 7

derivatives = [P] for k in range(1, 8): derivatives.append(sp.diff(derivatives[-1], z))

Point of evaluation

z0 = 3*sp.I

Compute the sum S = sum_{k=0}^{7} binom(7,k) * P^{(k)}(z0) * i^{7-k}

S = 0 for k in range(8): Pk = derivatives[k].subs(z, z0) binom = sp.binomial(7, k) power_i = sp.I**(7 - k) S += binom * Pk * power_i

Compute the integral I = 2pii / 7! * S

I = 2sp.pisp.I / sp.factorial(7) * S

Simplify

I_simplified = sp.simplify(I)

Get real and imaginary parts as floats with high precision

I_eval = I_simplified.evalf(30) a = I_eval.real b = I_eval.imag

print("I =", I_eval) print("a =", a) print("b =", b) print("10a =", 10a) print("floor(10a) =", int(10a // 1)) # integer division to get floor for positive? Need careful floor for negative. print("floor(b) =", int(b // 1)) print("floor(10a) + floor(b) =", int(10a // 1) + int(b // 1))

More precise floor: use math.floor

import math print("\nUsing math.floor:") print("floor(10a) =", math.floor(10a)) print("floor(b) =", math.floor(b)) print("Sum =", math.floor(10*a) + math.floor(b)) [/PYTHON]

[THOUGHT] The error occurred because I_eval is a symbolic expression, not a numeric complex number. I need to extract the real and imaginary parts properly using sp.re and sp.im and then evaluate numerically. Let me fix the code to get the correct numerical values for a and b.

import sympy as sp
import math

# Define variable
z = sp.symbols('z', complex=True)

# Define polynomial P(z) with complex coefficients
coeffs = [
    1,                                   # x^7
    -14*sp.I,                            # x^6
    (85 - 60*sp.I),                      # x^5
    -(240 + 210*sp.I),                    # x^4
    (420 - 840*sp.I),                     # x^3
    -(560 + 1680*sp.I),                   # x^2
    (1344 - 2688*sp.I),                   # x^1
    3840                                 # constant
]

# Build polynomial
P = sum(c * z**(7 - i) for i, c in enumerate(coeffs))

# Compute derivatives up to 7
derivatives = [P]
for k in range(1, 8):
    derivatives.append(sp.diff(derivatives[-1], z))

# Point of evaluation
z0 = 3*sp.I

# Compute the sum S = sum_{k=0}^{7} binom(7,k) * P^{(k)}(z0) * i^{7-k}
S = 0
for k in range(8):
    Pk = derivatives[k].subs(z, z0)
    binom = sp.binomial(7, k)
    power_i = sp.I**(7 - k)
    S += binom * Pk * power_i

# Compute the integral I = 2*pi*i / 7! * S
I = 2*sp.pi*sp.I / sp.factorial(7) * S

# Simplify
I_simplified = sp.simplify(I)

# Extract real and imaginary parts symbolically and evaluate numerically
a_exact = sp.re(I_simplified)
b_exact = sp.im(I_simplified)

a = a_exact.evalf(30)
b = b_exact.evalf(30)

print("I =", I_simplified.evalf(30))
print("a =", a)
print("b =", b)
print("10*a =", 10*a)
print("floor(10*a) =", math.floor(10*a))
print("floor(b) =", math.floor(b))
print("Sum =", math.floor(10*a) + math.floor(b))

[THOUGHT] From the Python output, we have a = 12.0278118737437798272569775246 and b = 40.9828241643297075084210073738. Then 10a = 120.278118737437798272569775246, so floor(10a) = 120. floor(b) = 40. Therefore, floor(10a) + floor(b) = 160.

\boxed{160}