Question

If $z = x - iy$ and ${z^{\dfrac{1}{3}}} = p + iq$, then $\dfrac{{\dfrac{x}{p} + \dfrac{y}{q}}}{{{p^2} + {q^2}}}$ is equal to
(A) $1$
(B) $- 1$
(C) $2$
(D) $- 2$

Answer 1

Here $z$ is a complex number. We will first simplify the equation ${z^{\dfrac{1}{3}}} = p + iq$ by taking a cube of both sides. Then we will equate this obtained value of $z$ in terms of $p$ and $q$ with the given $z = x - iy$ to obtain an equation. Solving this equation, we will find the value of $x$ and $y$ and then the value of $\dfrac{x}{p}$ and $\dfrac{y}{p}$. We will put these values in $\dfrac{{\dfrac{x}{p} + \dfrac{y}{q}}}{{{p^2} + {q^2}}}$ and then simplify it to find the result.

Complete step by step answer:
Given, ${z^{\dfrac{1}{3}}} = p + iq$
Taking cube of both the sides, we get
$\Rightarrow {\left( {{z^{\dfrac{1}{3}}}} \right)^3} = {\left( {p + iq} \right)^3}$
On simplifying we get
$\Rightarrow z = {\left( {p + iq} \right)^3}$
As ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b)$, using this we get
$\Rightarrow z = {p^3} + {i^3}{q^3} + 3ipq\left( {p + iq} \right)$
On simplifying,
$\Rightarrow z = {p^3} + {i^3}{q^3} + 3i{p^2}q + 3{i^2}p{q^2}$
As ${i^2} = - 1$ and ${i^3} = - i$. Using this we get
$\Rightarrow z = {p^3} + \left( { - i} \right) \times {q^3} + 3i{p^2}q + 3 \times \left( { - 1} \right) \times p{q^2}$
On solving,
$\Rightarrow z = {p^3} - i{q^3} + 3i{p^2}q - 3p{q^2}$
On rearranging we get
$\Rightarrow z = {p^3} - 3p{q^2} + 3i{p^2}q - i{q^3}$
Taking $i$ common from last two terms we get
$\Rightarrow z = \left( {{p^3} - 3p{q^2}} \right) + i\left( {3{p^2}q - {q^3}} \right)$
Now, putting $z = x - iy$ as given in the question
$\Rightarrow x - iy = \left( {{p^3} - 3p{q^2}} \right) + i\left( {3{p^2}q - {q^3}} \right)$
Here, $x$ is known as the real part of $z$ and $y$ is known as the imaginary part of $z$.
Now, we will compare the left-hand side of the equation with the right-hand side.
On comparing the real part and imaginary part we get,
\Rightarrow $$$$x = \left( {{p^3} - 3p{q^2}} \right) and $y = - \left( {3{p^2}q - {q^3}} \right)$
Now dividing $x$ by $p$ and $y$ by $q$, we get
\Rightarrow $$$$\dfrac{x}{p} = \dfrac{{\left( {{p^3} - 3p{q^2}} \right)}}{p} and $\dfrac{y}{q} = \dfrac{{ - \left( {3{p^2}q - {q^3}} \right)}}{q}$
On simplifying, we get
\Rightarrow $$$$\dfrac{x}{p} = {p^2} - 3{q^2} and $\dfrac{y}{q} = {q^2} - 3{p^2}$
On adding $\dfrac{x}{p}$ and $\dfrac{y}{q}$, we get
$\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = {p^2} - 3{q^2} + {q^2} - 3{p^2}$
On solving we get
$\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = \left( { - 2{q^2} - 2{p^2}} \right)$
Taking $( - 2)$ common from right-hand side, we get
$\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = - 2\left( {{p^2} + {q^2}} \right)$
Dividing the above equation by $\left( {{p^2} + {q^2}} \right)$, we get
$\Rightarrow \dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = \dfrac{{ - 2\left( {{p^2} + {q^2}} \right)}}{{\left( {{p^2} + {q^2}} \right)}}$
On simplification, we get
$\Rightarrow \dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = - 2$
Therefore, $\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}}$ is equal to $- 2$.
So, the correct answer is “Option D”.

Note:
Here it is given that $z = x - iy$ , that’s why we have taken cube of both the sides to simplify ${z^{\dfrac{1}{3}}} = p + iq$ accordingly in a way to get an equation in $z$, so that we can equate both the given equation. One thing to note is that we have to keep in mind that since it is given $z = x - iy$, we have to simplify the other equation ${z^{\dfrac{1}{3}}} = p + iq$ in such a way that it helps to solve the problem. For example, if we would have taken squares of both sides then it would have complicated the problem.