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Question: If \[z = x - iy\] and \[{z^{\dfrac{1}{3}}} = p + iq\], then \[\dfrac{{\dfrac{x}{p} + \dfrac{y}{q}}}{...

If z=xiyz = x - iy and z13=p+iq{z^{\dfrac{1}{3}}} = p + iq, then xp+yqp2+q2\dfrac{{\dfrac{x}{p} + \dfrac{y}{q}}}{{{p^2} + {q^2}}} is equal to
(A) 11
(B) 1 - 1
(C) 22
(D) 2 - 2

Explanation

Solution

Here zz is a complex number. We will first simplify the equation z13=p+iq{z^{\dfrac{1}{3}}} = p + iq by taking a cube of both sides. Then we will equate this obtained value of zz in terms of pp and qq with the given z=xiyz = x - iy to obtain an equation. Solving this equation, we will find the value of xx and yy and then the value of xp\dfrac{x}{p} and yp\dfrac{y}{p}. We will put these values in xp+yqp2+q2\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, z13=p+iq{z^{\dfrac{1}{3}}} = p + iq
Taking cube of both the sides, we get
(z13)3=(p+iq)3\Rightarrow {\left( {{z^{\dfrac{1}{3}}}} \right)^3} = {\left( {p + iq} \right)^3}
On simplifying we get
z=(p+iq)3\Rightarrow z = {\left( {p + iq} \right)^3}
As (a+b)3=a3+b3+3ab(a+b){\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b), using this we get
z=p3+i3q3+3ipq(p+iq)\Rightarrow z = {p^3} + {i^3}{q^3} + 3ipq\left( {p + iq} \right)
On simplifying,
z=p3+i3q3+3ip2q+3i2pq2\Rightarrow z = {p^3} + {i^3}{q^3} + 3i{p^2}q + 3{i^2}p{q^2}
As i2=1{i^2} = - 1 and i3=i{i^3} = - i. Using this we get
z=p3+(i)×q3+3ip2q+3×(1)×pq2\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,
z=p3iq3+3ip2q3pq2\Rightarrow z = {p^3} - i{q^3} + 3i{p^2}q - 3p{q^2}
On rearranging we get
z=p33pq2+3ip2qiq3\Rightarrow z = {p^3} - 3p{q^2} + 3i{p^2}q - i{q^3}
Taking ii common from last two terms we get
z=(p33pq2)+i(3p2qq3)\Rightarrow z = \left( {{p^3} - 3p{q^2}} \right) + i\left( {3{p^2}q - {q^3}} \right)
Now, putting z=xiyz = x - iy as given in the question
xiy=(p33pq2)+i(3p2qq3)\Rightarrow x - iy = \left( {{p^3} - 3p{q^2}} \right) + i\left( {3{p^2}q - {q^3}} \right)
Here, xx is known as the real part of zz and yy is known as the imaginary part of zz.
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 $$$$x = \left( {{p^3} - 3p{q^2}} \right) and y=(3p2qq3)y = - \left( {3{p^2}q - {q^3}} \right)
Now dividing xx by pp and yy by qq, we get
\Rightarrow $$$$\dfrac{x}{p} = \dfrac{{\left( {{p^3} - 3p{q^2}} \right)}}{p} and yq=(3p2qq3)q\dfrac{y}{q} = \dfrac{{ - \left( {3{p^2}q - {q^3}} \right)}}{q}
On simplifying, we get
\Rightarrow $$$$\dfrac{x}{p} = {p^2} - 3{q^2} and yq=q23p2\dfrac{y}{q} = {q^2} - 3{p^2}
On adding xp\dfrac{x}{p} and yq\dfrac{y}{q}, we get
xp+yq=p23q2+q23p2\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = {p^2} - 3{q^2} + {q^2} - 3{p^2}
On solving we get
xp+yq=(2q22p2)\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = \left( { - 2{q^2} - 2{p^2}} \right)
Taking (2)( - 2) common from right-hand side, we get
xp+yq=2(p2+q2)\Rightarrow \dfrac{x}{p} + \dfrac{y}{q} = - 2\left( {{p^2} + {q^2}} \right)
Dividing the above equation by (p2+q2)\left( {{p^2} + {q^2}} \right), we get
(xp+yq)(p2+q2)=2(p2+q2)(p2+q2)\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
(xp+yq)(p2+q2)=2\Rightarrow \dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = - 2
Therefore, (xp+yq)(p2+q2)\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} is equal to 2 - 2.
So, the correct answer is “Option D”.

Note:
Here it is given that z=xiyz = x - iy , that’s why we have taken cube of both the sides to simplify z13=p+iq{z^{\dfrac{1}{3}}} = p + iq accordingly in a way to get an equation in zz, 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=xiyz = x - iy, we have to simplify the other equation z13=p+iq{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.