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

Hint: 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.