Question
Question: If \[z = x - iy\] and \[{z^{\dfrac{1}{3}}} = p + iq\], then \[\dfrac{{\dfrac{x}{p} + \dfrac{y}{q}}}{...
If z=x−iy and z31=p+iq, then p2+q2px+qy is equal to
(A) 1
(B) −1
(C) 2
(D) −2
Solution
Here z is a complex number. We will first simplify the equation z31=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 px and py. We will put these values in p2+q2px+qy and then simplify it to find the result.
Complete step by step answer:
Given, z31=p+iq
Taking cube of both the sides, we get
⇒z313=(p+iq)3
On simplifying we get
⇒z=(p+iq)3
As (a+b)3=a3+b3+3ab(a+b), using this we get
⇒z=p3+i3q3+3ipq(p+iq)
On simplifying,
⇒z=p3+i3q3+3ip2q+3i2pq2
As i2=−1 and i3=−i. Using this we get
⇒z=p3+(−i)×q3+3ip2q+3×(−1)×pq2
On solving,
⇒z=p3−iq3+3ip2q−3pq2
On rearranging we get
⇒z=p3−3pq2+3ip2q−iq3
Taking i common from last two terms we get
⇒z=(p3−3pq2)+i(3p2q−q3)
Now, putting z=x−iy as given in the question
⇒x−iy=(p3−3pq2)+i(3p2q−q3)
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 $$$$x = \left( {{p^3} - 3p{q^2}} \right) and y=−(3p2q−q3)
Now dividing x by p and y by q, we get
\Rightarrow $$$$\dfrac{x}{p} = \dfrac{{\left( {{p^3} - 3p{q^2}} \right)}}{p} and qy=q−(3p2q−q3)
On simplifying, we get
\Rightarrow $$$$\dfrac{x}{p} = {p^2} - 3{q^2} and qy=q2−3p2
On adding px and qy, we get
⇒px+qy=p2−3q2+q2−3p2
On solving we get
⇒px+qy=(−2q2−2p2)
Taking (−2) common from right-hand side, we get
⇒px+qy=−2(p2+q2)
Dividing the above equation by (p2+q2), we get
⇒(p2+q2)(px+qy)=(p2+q2)−2(p2+q2)
On simplification, we get
⇒(p2+q2)(px+qy)=−2
Therefore, (p2+q2)(px+qy) 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 z31=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 z31=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.