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If $z = x + iy$ is a complex number such that ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$, then the value of $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right) =$A. -1B. -2C. 0D. 2

Last updated date: 19th Jun 2024
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Hint: according to the question we have to find the value of $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$when $z = x + iy$ is a complex number such that ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$
So, first of all we have to taking cube both side of the given expression ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$and put the conjugate of $z$ that is $\overline z = x - iy$
Formula used for the cube of $\left( {a + b} \right)$ that is mentioned below.

Formula used:
${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b).............................(A)$
Now, we have to compare both real and imaginary roots of $x - iy$and the expression obtained after taking the cube of $a + ib$to get the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$
Now, we have to put the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$in the given expression $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$to get the desired value.

Step 1: First of all we have to taking cube both side of the given expression ${\left( {\overline z } \right)^{\dfrac{1}{3}}} = a + ib$
$\Rightarrow \overline z = {(a + ib)^3}$
Now, use the formula of cube (A) that is mentioned in the solution hint.
$\Rightarrow (x - iy) = {a^3} + {\left( {ib} \right)^3} + 3\left( a \right)\left( {ib} \right)\left( {a + ib} \right) \\ \Rightarrow (x - iy) = {a^3} + {i^3}{b^3} + 3{a^2}\left( {ib} \right) + 3a{\left( {ib} \right)^2} \\$
As we know that ${i^2} = - 1$and ${i^3} = - i$
$\Rightarrow (x - iy) = {a^3} - i{b^3} + 3{a^2}\left( {ib} \right) - 3a{b^2} \\ \Rightarrow (x - iy) = {a^3} - 3a{b^2} + i\left( {3{a^2}b - {b^3}} \right).......................(1) \\$
Step 3: Now, we have to compare the both real and imaginary values of the expression (1) as obtained in the solution step 2.
$\Rightarrow x = \left( {{a^3} - 3a{b^2}} \right)$and, $y = - \left( {\left( {3{a^2}b - {b^3}} \right)} \right)$
$\Rightarrow \dfrac{x}{a} = \left( {{a^2} - 3{b^2}} \right)$and, $\dfrac{y}{b} = \left( {\left( { - 3{a^2} + {b^2}} \right)} \right)...............................(2)$
Step 4: Now, we have to the values of $\dfrac{x}{a}$and $\dfrac{y}{b}$from the expression (2) in the given expression$\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right)$
$\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( {\left( {{a^2} - 3{b^2}} \right) + \left( { - 3{a^2} + {b^2}} \right)} \right) \\ \Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( {{a^2} - 3{a^2} + {b^2} - 3{b^2}} \right) \\$
Now, solving the expression as obtained just above,
$\Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left( { - 2{b^2} - 2{a^2}} \right) \\ \Rightarrow \dfrac{1}{{{a^2} + {b^2}}}\left\{ { - 2\left( {{a^2} + {b^2}} \right)} \right\} \\$
On eliminating the terms which can be eliminated,
$\Rightarrow - 2$

Hence, we have obtained the value of $\dfrac{1}{{{a^2} + {b^2}}}\left( {\dfrac{x}{a} + \dfrac{y}{b}} \right) = - 2$. Therefore option (B) is correct.

Note:
It is necessary that we have to find the cube of the expression given in the question then we can compare the real and imaginary roots with the conjugate of z which is $\overline z = x - iy$ and where z is $= x + iy$.
It is necessary that we have to find the value of $\dfrac{x}{a}$and $\dfrac{y}{b}$with the help of comparing the expression of terms x and y.