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**Hint:**We are given an equation of fourth degree and its roots are complex. We will first write the roots of the equation i. e., $\left( {x - a - ib} \right),\left( {x - a + ib} \right),\left( {x - b - ia} \right),\left( {x - b + ia} \right)$. Then, we will form an equation using the given roots and on simplification, we will compare it with the given equation. And, by comparison, we will find the values of a and b. After that we will check the availability of every option.

**Complete step-by-step answer:**We know that if an equation has real coefficients and imaginary roots, the roots always occur in pairs and the same situation is given in the question.

Now, let f (x) = 0 be the equation formed by the given roots $a \pm ib$ and $b \pm ia$.

Since, $a \pm ib$ and $b \pm ia$ are the roots of the f (x) = 0, then

$\left( {x - a - ib} \right),\left( {x - a + ib} \right),\left( {x - b - ia} \right),\left( {x - b + ia} \right)$ are the factors of the polynomial f (x).

Hence, the desired equation will be:

f (x) = $\left[ {\left( {x - a - ib} \right)\left( {x - a + ib} \right)\left( {x - b - ia} \right)\left( {x - b + ia} \right)} \right] = 0$

Now, when we have the equation, we will simplify this equation in terms of powers of x to make it easy for comparing the given equation with this.

Using the algebraic identity: $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$in the above equation, we get

$ \Rightarrow $f (x) = $\left[ {{{\left( {x - a} \right)}^2} - {{\left( {ib} \right)}^2}} \right]\left[ {{{\left( {x - b} \right)}^2} - {{\left( {ia} \right)}^2}} \right] = 0$

As we know that ${i^2} = - 1$, putting the value in the above equation, we get

$ \Rightarrow f(x) = \left[ {{{\left( {x - a} \right)}^2} + {b^2}} \right]\left[ {{{\left( {x - b} \right)}^2} + {a^2}} \right] = 0$

Opening the brackets using the identity: ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we get

$ \Rightarrow f(x) = \left[ {{x^2} - 2ax + {a^2} + {b^2}} \right]\left[ {{x^2} - 2bx + {b^2} + {a^2}} \right] = 0$

On further simplification, we get

$f(x) = \left[ {{x^4} - 2a{x^3} + \left( {{a^2} + {b^2}} \right){x^2} + \left( {{a^2} + {b^2}} \right){x^2} - 2ax\left( {{a^2} + {b^2}} \right) + {{\left( {{a^2} + {b^2}} \right)}^2} - 2b{x^3} + 4ab{x^2} - 2bx\left( {{a^2} + {b^2}} \right)} \right] = 0$

Combining the terms with the same powers of x and then simplifying the terms, we get

$ \Rightarrow f(x) = \left[ {{x^4} - 2{x^3}\left( {a + b} \right) + 2{x^2}{{\left( {a + b} \right)}^2} - 2\left( {{a^2} + {b^2}} \right)\left( {a + b} \right)x + {{\left( {{a^2} + {b^2}} \right)}^2}} \right] = 0$

Now, we get an equation which is in only form of coefficient of different powers of x. hence, comparing this equation $ \Rightarrow f(x) = \left[ {{x^4} - 2{x^3}\left( {a + b} \right) + 2{x^2}{{\left( {a + b} \right)}^2} - 2\left( {{a^2} + {b^2}} \right)\left( {a + b} \right)x + {{\left( {{a^2} + {b^2}} \right)}^2}} \right] = 0$ with the given equation ${x^4} - 10{x^3} + 50{x^2} - 130x + 169 = 0$, we get

$

2(a + b) = 10 \\

2{(a + b)^2} = 50 \\

2({a^2} + {b^2})(a + b) = 130 \\

{({a^2} + {b^2})^2} = 169 \\

$

Checking for every option:

Option(A): (3, 2). If it is the value of (a, b) then it must satisfy the above mentioned condition.

$\therefore $ for 2(a + b) = 10 $ \Rightarrow 2\left( {3 + 2} \right) = 2\left( 5 \right) = 10$

Therefore, option (A) is correct.

Option(B): (2, 1). Again, it should satisfy the conditions.

For $2\left( {a + b} \right) = 10 \Rightarrow 2(2 + 1) = 2(3) = 6 \ne 10$

Hence, option(B) is incorrect.

Option(C): (-3, 2).

For $2(a + b) = 10 \Rightarrow 2( - 3 + 2) = 2( - 1) = - 2 \ne 10$

Hence, option(C) is incorrect.

Option(D): (-3, -2).

For $2(a + b) = 10 \Rightarrow 2( - 3 - 2) = 2( - 5) = - 10 \ne 10$

Therefore, option(D) is incorrect.

**Hence, option A is correct.**

**Note:**In such problems, it’s lengthy while solving the equations to reduce it into a simpler form. Such questions where roots are given and variables are required to find, you can also solve them by using the formula to form an equation of 4th degree. In this case you can directly solve by applying the formula 2(a+b) for the sum of the roots of a 4th degree equation.

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