<Question>
Find the sum of all complex solutions of the equation
\[\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.\]
</Question>

<think>
<Plan_1>
Okay, so I need to find the sum of all complex solutions of this equation:

\[\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.\]

Hmm, that looks a bit complicated. Let me start by understanding what the equation is asking for. It's a rational equation on the left side with four terms, each having a linear numerator and a quadratic denominator. The right side is a linear function, 2010x - 4. The problem is to find all complex solutions x and then sum them up.

Since the equation involves fractions, I should check for any restrictions on x. The denominators x² - 1, x² - 2, x² - 3, x² - 4 cannot be zero. Therefore, x ≠ ±1, ±√2, ±√3, ±2. These are the values that would make the denominators zero, so they are excluded from the domain.

Now, to solve the equation, maybe I can clear the denominators by multiplying both sides by the common denominator. Let's see. The denominators are (x² - 1), (x² - 2), (x² - 3), (x² - 4). So the common denominator would be the product of all four: (x² - 1)(x² - 2)(x² - 3)(x² - 4). That seems like a lot, but perhaps there's a smarter way.

Alternatively, maybe there's a substitution that can simplify the equation. Let me think.

<prompt_1.1> Consider possible substitutions to simplify the equation and plan the next steps. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The left side has terms like 1/(x² - 1), 2/(x² - 2), etc. Each term has a numerator equal to the number before the denominator, which is 1, 2, 3, 4. Maybe if I let y = x², then each term becomes k/(y - k) where k is 1, 2, 3, 4. Let's try that substitution.

Let y = x². Then the equation becomes:

1/(y - 1) + 2/(y - 2) + 3/(y - 3) + 4/(y - 4) = 2010x - 4.

Hmm, but now we have a mix of y and x. Since y = x², maybe this substitution isn't directly helpful because we still have x in the right side. However, perhaps if we can express x in terms of y? But x = ±√y, which might complicate things. Alternatively, maybe there's another way.

Wait, but the equation is in terms of x, and after substitution, it's still a relation between x and y. Maybe this substitution isn't helpful. Let's think again.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Alternatively, maybe consider combining the fractions on the left side. Let's see.

<prompt_2.1> Attempt to combine the fractions on the left side and simplify the equation. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
The left side is the sum of four fractions. To combine them, we need a common denominator, which is the product of all four denominators. Let's denote D = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Then, each term can be written as:

1/(x² - 1) = [ (x² - 2)(x² - 3)(x² - 4) ] / D,

and similarly for the others. So multiplying each term by D and then summing up would give:

[ (x² - 2)(x² - 3)(x² - 4) ] + 2[ (x² - 1)(x² - 3)(x² - 4) ] + 3[ (x² - 1)(x² - 2)(x² - 4) ] + 4[ (x² - 1)(x² - 2)(x² - 3) ] = (2010x - 4)D.

But expanding all of that seems really tedious. Maybe there's a pattern or a simplification here.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me think. Let's denote t = x². Then each denominator is t - k for k = 1,2,3,4. So the left side becomes:

<prompt_3.1> Re-evaluate the substitution idea and consider if there's a pattern or simplification. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
1/(t - 1) + 2/(t - 2) + 3/(t - 3) + 4/(t - 4).

So, maybe if I can compute this sum in terms of t, then set it equal to 2010x - 4, and then relate t to x. But since t = x², this might not directly help because x is still present on the right side.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Alternatively, perhaps there is a telescoping nature here. Let me check.

<prompt_4.1> Explore the possibility of a telescoping series in the left side. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
If I consider the left side:

1/(x² - 1) + 2/(x² - 2) + 3/(x² - 3) + 4/(x² - 4).

Is there a way to write each term as a difference of fractions that cancels out with adjacent terms? For example, sometimes when you have a sum of fractions like A/(x - a) + B/(x - b), you can write it as (A/(x - a) - A/(x - b)) + something. But here, the denominators are quadratic. Wait, but x² - k can be factored as (x - √k)(x + √k). So maybe partial fractions?

But the numerators here are 1, 2, 3, 4, which are linear in the term number. Let me see:

First term: 1/(x² - 1) = 1/[(x - 1)(x + 1)]. Similarly, the second term is 2/[(x - √2)(x + √2)], and so on. Maybe each term can be decomposed into partial fractions. Let's try that.

For the first term: 1/[(x - 1)(x + 1)] = A/(x - 1) + B/(x + 1). Solving for A and B:

1 = A(x + 1) + B(x - 1). Let x = 1: 1 = A(2) => A = 1/2. Let x = -1: 1 = B(-2) => B = -1/2. So 1/(x² - 1) = (1/2)[1/(x - 1) - 1/(x + 1)].

Similarly, the second term: 2/[(x - √2)(x + √2)] = C/(x - √2) + D/(x + √2). Multiply both sides by (x - √2)(x + √2):

2 = C(x + √2) + D(x - √2). Let x = √2: 2 = C(2√2) => C = 2/(2√2) = 1/√2. Let x = -√2: 2 = D(-2√2) => D = -2/(2√2) = -1/√2. Therefore, 2/(x² - 2) = (1/√2)[1/(x - √2) - 1/(x + √2)].

Similarly, third term: 3/(x² - 3) = 3/[(x - √3)(x + √3)] = E/(x - √3) + F/(x + √3). Solving:

3 = E(x + √3) + F(x - √3). Let x = √3: 3 = E(2√3) => E = 3/(2√3) = √3/2. Let x = -√3: 3 = F(-2√3) => F = -3/(2√3) = -√3/2. Therefore, 3/(x² - 3) = (√3/2)[1/(x - √3) - 1/(x + √3)].

Fourth term: 4/(x² - 4) = 4/[(x - 2)(x + 2)] = G/(x - 2) + H/(x + 2). Solving:

4 = G(x + 2) + H(x - 2). Let x = 2: 4 = G(4) => G = 1. Let x = -2: 4 = H(-4) => H = -1. So 4/(x² - 4) = [1/(x - 2) - 1/(x + 2)].

So, putting all these partial fractions together, the left side becomes:
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
(1/2)[1/(x - 1) - 1/(x + 1)] + (1/√2)[1/(x - √2) - 1/(x + √2)] + (√3/2)[1/(x - √3) - 1/(x + √3)] + [1/(x - 2) - 1/(x + 2)] = 2010x - 4.

Hmm, so each term is of the form A[1/(x - a) - 1/(x + a)], where A and a are constants. Let's see if we can combine these terms.

<prompt_5.1> Combine the transformed terms and simplify the equation. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Let me write all the coefficients:

For 1/(x - 1): 1/2.

For 1/(x + 1): -1/2.

For 1/(x - √2): 1/√2.

For 1/(x + √2): -1/√2.

For 1/(x - √3): √3/2.

For 1/(x + √3): -√3/2.

For 1/(x - 2): 1.

For 1/(x + 2): -1.

So combining all these, the left side can be written as:

[1/2]/(x - 1) - [1/2]/(x + 1) + [1/√2]/(x - √2) - [1/√2]/(x + √2) + [√3/2]/(x - √3) - [√3/2]/(x + √3) + 1/(x - 2) - 1/(x + 2) = 2010x - 4.

Hmm, this seems complicated, but maybe if we consider the entire left side as a sum of simple fractions, there's a way to combine them. Alternatively, perhaps we can note that each pair like [1/(x - a) - 1/(x + a)] can be combined into 2a/(x² - a²). Wait, but that's how we started. So maybe this approach isn't helpful.

Alternatively, perhaps the left side can be expressed as a derivative or something else. Wait, but maybe if we consider the sum of all these terms, we can think of it as a sum over residues or something. Wait, but this is a complex analysis thought, and the problem is about complex solutions. Maybe there's a connection.

Alternatively, perhaps considering that the left side is a sum of terms each of which has a simple pole at x = ±√k, and the right side is a linear function. But I'm not sure how that helps. Maybe if we consider the equation as a rational function equal to a linear function, then the equation can be written as a single rational function minus 2010x + 4 equals zero. Then, the sum of roots can be found via Vieta's formula if we can write the equation as a polynomial.

But to do that, we need to clear the denominators. Let's try that. Let's denote D = (x² - 1)(x² - 2)(x² - 3)(x² - 4). Then, multiplying both sides by D gives:
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
N = (x² - 2)(x² - 3)(x² - 4) + 2(x² - 1)(x² - 3)(x² - 4) + 3(x² - 1)(x² - 2)(x² - 4) + 4(x² - 1)(x² - 2)(x² - 3).

So expanding N and then moving everything to one side would give a polynomial equation. Then, the sum of all complex solutions is equal to minus the coefficient of x^{2n - 1} divided by the coefficient of x^{2n}, where 2n is the degree of the polynomial.

But expanding N seems really tedious. Let me see if there's a smarter way.

<prompt_6.1> Consider if there's a smarter way to find the sum of roots without expanding. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Alternatively, notice that the original equation is a sum of terms each with denominator x² - k, which are quadratics. When we multiply through by D, which is the product of all four denominators, each term in N will be the product of three denominators. For example, the first term is (x² - 2)(x² - 3)(x² - 4), and so on. So N is a polynomial of degree 6, since each term in N is a product of three quadratics, which is degree 6. Then, setting N - (2010x - 4)D = 0, which would be a degree 6 polynomial equation. Therefore, the sum of all roots (complex included) is given by Vieta's formula: the coefficient of x^5 divided by the coefficient of x^6, with a sign change.

But computing the coefficients of x^6 and x^5 in N and D might be feasible without expanding everything.

First, let's find the degree of N. As mentioned, each term in N is a product of three quadratics, so each term is a degree 6 polynomial. Therefore, N is a degree 6 polynomial. Similarly, D is a product of four quadratics, so D is degree 8. Therefore, when we write N - (2010x - 4)D = 0, the term (2010x - 4)D is a polynomial of degree 9 (since 2010x is degree 1 times D which is degree 8). Therefore, the equation is a polynomial of degree 9. However, the original equation after multiplying both sides by D would result in a degree 9 polynomial, but perhaps some terms cancel out.

Wait, but let's check:

Left side after multiplying by D: N, which is degree 6.

Right side after multiplying by D: (2010x - 4)D, which is degree 8 + 1 = 9. Therefore, moving everything to left side: N - (2010x - 4)D is a polynomial of degree 9. Therefore, the equation is a degree 9 polynomial. Therefore, the sum of all roots (complex included) is equal to minus the coefficient of x^8 divided by the coefficient of x^9, by Vieta's formula.

Therefore, if we can compute the coefficients of x^9 and x^8 in the polynomial N - (2010x - 4)D, then we can find the sum of all roots.

So let's compute these coefficients.

First, let's compute the coefficient of x^9 in (2010x - 4)D. Since D is (x² - 1)(x² - 2)(x² - 3)(x² - 4), which is a product of four quadratics. The leading term of D is x^8, so D = x^8 + lower degree terms. Therefore, (2010x - 4)D = 2010x * D - 4D. The leading term of 2010x * D is 2010x * x^8 = 2010x^9. The leading term of -4D is -4x^8. Therefore, the coefficient of x^9 in (2010x - 4)D is 2010, and the coefficient of x^8 is -4.

Now, the coefficient of x^9 in N is zero, because N is a product of three quadratics, each of which is degree 2, so N is degree 6. Therefore, in the polynomial N - (2010x - 4)D, the coefficient of x^9 is 0 - 2010 = -2010. The coefficient of x^8 is 0 - (-4) = 4.

Wait, hold on. Wait, N is degree 6, so when we subtract (2010x - 4)D, which is degree 9, the resulting polynomial is degree 9. The coefficient of x^9 comes only from the (2010x - 4)D term, since N doesn't have an x^9 term. So coefficient of x^9 in N - (2010x -4)D is 0 - 2010x^9 term, which is -2010. Similarly, the coefficient of x^8 in N is 0 (since N is degree 6), and in (2010x -4)D, the x^8 term comes from -4D. The leading term of D is x^8, so -4D has a term -4x^8. Therefore, the coefficient of x^8 in (2010x -4)D is -4. Therefore, in N - (2010x -4)D, the coefficient of x^8 is 0 - (-4) = 4.

Therefore, the polynomial equation is of the form:

-2010x^9 + 4x^8 + ... (lower degree terms) = 0.

Therefore, according to Vieta's formula, the sum of all roots is equal to - (coefficient of x^8) / (coefficient of x^9) = - (4)/(-2010) = 4/2010 = 2/1005.

Wait, 4 divided by 2010 is 2/1005. So the sum of all complex solutions is 2/1005.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But hold on, this seems too straightforward. Let me verify.

<prompt_7.1> Verify the calculated sum of roots. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Because when we clear denominators, the polynomial is degree 9, and Vieta's formula gives the sum of roots as - (coefficient of x^8)/(coefficient of x^9). The coefficient of x^9 is -2010, and coefficient of x^8 is 4, so sum is - (4)/(-2010) = 4/2010 = 2/1005.

But let me make sure there are no mistakes here. Let's retrace the steps.

1. Original equation: sum of four terms = 2010x -4.

2. Multiply both sides by D = (x² -1)(x² -2)(x² -3)(x² -4).

3. Left side becomes N, which is a product of three denominators each time, so degree 6.

4. Right side becomes (2010x -4)D, which is degree 9.

5. Therefore, moving everything to left side: N - (2010x -4)D is a degree 9 polynomial.

6. The coefficient of x^9 in N is 0, in (2010x -4)D it's -2010x^9 (from 2010x * x^8 term in D), so coefficient of x^9 in the equation is 0 - (-2010) = +2010? Wait, wait, no. Wait, when we have N - (2010x -4)D = 0, so the coefficient of x^9 is coefficient of x^9 in N minus coefficient of x^9 in (2010x -4)D. Since N has no x^9 term, it's 0 - (2010) = -2010. Therefore, the coefficient of x^9 in the equation is -2010. Then the coefficient of x^8 is 0 - (-4) = +4. Therefore, the polynomial is -2010x^9 + 4x^8 + ... = 0.

Therefore, the sum of roots is - (coefficient of x^8) / (coefficient of x^9) = - (4)/(-2010) = 4/2010 = 2/1005.

Simplify 2/1005: 1005 = 5*201 = 5*3*67. So 2 and 1005 are coprime. So 2/1005 is the simplified form.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
But let me check if there are any other terms that might contribute to x^9 or x^8.

<prompt_8.1> Re-evaluate the polynomial degree and confirm the sum of roots. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
Wait, N is degree 6, so when subtracting (2010x -4)D, which is degree 9, the resulting polynomial is indeed degree 9. The x^9 term only comes from the (2010x -4)D term, since N doesn't have an x^9 term. Similarly, the x^8 term in (2010x -4)D comes from -4D. Since D is x^8 + ... , so -4D has a term -4x^8. Therefore, when subtracting (2010x -4)D, the x^8 term becomes +4x^8. Therefore, in the polynomial equation, the x^8 coefficient is +4, and x^9 coefficient is -2010. Therefore, the sum of roots is - (4)/(-2010) = 4/2010 = 2/1005.

Therefore, the answer should be 2/1005.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
But let me confirm once more. Let's consider the degrees:

<prompt_9.1> Confirm the solution through degree analysis. </prompt_9.1>
</Plan_9>

<Execution_9>
<execution_9.1>
D is degree 8, so (2010x -4)D is degree 9. N is degree 6. So when we write N - (2010x -4)D = 0, the highest degree term is x^9 with coefficient -2010. The next term is x^8 with coefficient 4. So yes, the polynomial is of degree 9, and Vieta's formula gives the sum of roots as - (coefficient of x^8)/(coefficient of x^9) = - (4)/(-2010) = 2/1005.

Therefore, the sum of all complex solutions is 2/1005.
</execution_9.1>
</Execution_9>

<Plan_10>
Based on execution_9.1:
But let me check if there's a miscalculation here. 2010x * D: D has leading term x^8, so 2010x * x^8 = 2010x^9.

<prompt_10.1> Verify the calculations for the polynomial coefficients. </prompt_10.1>

Therefore, the answer is 2/1005.

But wait, 1005 is 5*201, which is 5*3*67. So 2/1005 is in simplest terms. So the sum is 2/1005.

But let me check with a simpler case. Suppose instead of 2010x -4, the right side was a constant.

<prompt_10.2> Consider a simpler case to verify the approach. </prompt_10.2>

Therefore, applying the same logic here, the sum of roots is 2/1005.

But let me check another angle. Suppose we consider the original equation:

<prompt_10.3> Re-confirm the solution through substitution and symmetry. </prompt_10.3>
</Plan_10>

<Execution_10>
<execution_10.1>
So the coefficient of x^9 in (2010x -4)D is indeed -2010 (from -4D's x^8 term? Wait, no. Wait, (2010x -4)D = 2010x*D -4D. The term 2010x*D will have a leading term 2010x*(x^8) = 2010x^9, so coefficient 2010. The term -4D will have a leading term -4x^8. Therefore, the coefficient of x^9 in (2010x -4)D is 2010 (from 2010x*D) and the coefficient of x^8 is -4 (from -4D). Therefore, when we subtract (2010x -4)D from N, the coefficient of x^9 becomes 0 - 2010 = -2010, and the coefficient of x^8 becomes 0 - (-4) = +4. Therefore, the polynomial is -2010x^9 +4x^8 + ... =0. Therefore, the sum of roots is - (4)/(-2010) = 4/2010 = 2/1005.
</execution_10.1>

<execution_10.2>
Then, multiplying through by D would give a polynomial equation of degree 8 (from D) plus degree 0 (from the constant term times D), so the max degree is 8. Then sum of roots would be coefficient of x^7 over coefficient of x^8, etc. But in our case, since the right side is degree 1, when multiplied by D (degree 8), gives degree 9. So the logic holds.

Alternatively, perhaps we can test with specific values. Suppose x is a root, then 2010x -4 must equal the left side. If x is a complex number, but perhaps for certain x, the left side can be simplified.

But given the complexity, the Vieta's approach seems the most straightforward.

Therefore, after thorough analysis, the sum of all complex solutions is 2/1005.
</execution_10.2>

<execution_10.3>
Sum_{k=1 to 4} [1/(x² - k)] = 2010x -4.

If we consider that each term 1/(x² - k) is a function that is symmetric with respect to x and -x, except for the right side, which is linear in x. However, the right side 2010x -4 is not symmetric. Therefore, the equation is not symmetric with respect to x and -x. Therefore, the roots may not come in pairs that sum to zero. However, the problem asks for the sum of all complex solutions, which Vieta's formula gives as 2/1005.

But let me check with a simpler equation. Suppose instead of 2010x -4, the right side was a constant, say 0. Then, the equation would be:

Sum_{k=1 to 4} [1/(x² - k)] = 0.

Then, multiplying both sides by D, we get N = 0, so N is a degree 6 polynomial. Then, the sum of roots would be - (coefficient of x^5)/ (coefficient of x^6). But in that case, perhaps the sum would be different. However, in our problem, the right side is 2010x -4, which when multiplied by D gives a degree 9 polynomial. So the previous calculation holds.

Therefore, after thorough analysis, the sum of all complex solutions is 2/1005.
</execution_10.3>
</Execution_10>

<Plan_11>
Based on execution_10.1:
But let me check once more. Let's suppose that after expanding all terms, the coefficients of x^9 and x^8 are as we found. So the polynomial is:

-2010x^9 + 4x^8 + ... (lower terms) = 0.

Dividing both sides by -2010, we get:

<prompt_11.1> Verify the final polynomial and its root sum. </prompt_11.1>
</Plan_11>

<Execution_11>
<execution_11.1>
x^9 - (4/2010)x^8 + ... = 0.

But Vieta's formula states that the sum of roots is equal to the coefficient of x^8 divided by the coefficient of x^9, with a sign change. Wait, no. Wait, in general, for a polynomial:

a_n x^n + a_{n-1}x^{n-1} + ... + a_0 = 0,

the sum of roots is -a_{n-1}/a_n.

In our case, the polynomial is:

-2010x^9 + 4x^8 + ... = 0.

So here, a_9 = -2010, a_8 = 4. Therefore, the sum of roots is -a_8 / a_9 = -4 / (-2010) = 4/2010 = 2/1005.

Yes, that's correct. Therefore, the sum of all complex solutions is 2/1005.
</execution_11.1>
</Execution_11>

<Final_answer>
Therefore, the sum of all complex solutions of the given equation is $\boxed{\dfrac{2}{1005}}$.
</Final_answer>
</think>