Question

If $z = x - iy$ and ${z^{\dfrac{1}{3}}} = p + iq$ , then, what is the value of $\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}}$ ?
A) 1
B) -1
C) 2
D) -2

Answer 1

Hint: Iota, $i$ is a complex number. The square of iota results in -1, that is, ${i^2} = - 1$ . In the above question, take the cube of ${z^{\dfrac{1}{3}}}$ to evaluate the value of $z$ . Then, proceed further by comparing the real and imaginary parts of the complex number to obtain the desired value.

Complete step by step Solution:
It is given that: ${z^{\dfrac{1}{3}}} = p + iq$
Taking the cube,
$z = {\left( {p + iq} \right)^3}$
Opening the cube,
$z = {p^3} - i{q^3} + 3i{p^2}q - 3p{q^2}$
Separating the real and the imaginary part,
$z = {p^3} - 3p{q^2} + i(3{p^2}q - {q^3})$
It is also given that $z = x - iy$ .
Therefore,
$x - iy = {p^3} - 3p{q^2} + i(3{p^2}q - {q^3})$
Comparing the real and imaginary parts,
$x = {p^3} - 3p{q^2}$ and $y = - 3{p^2}q + {q^3}$
Now,
$\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = \dfrac{{\dfrac{{{p^3} - 3p{q^2}}}{p} + \dfrac{{ - 3{p^2}q + {q^3}}}{q}}}{{\left( {{p^2} + {q^2}} \right)}}$
Simplifying further,
$\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = \dfrac{{{p^2} - 3{q^2} - 3{p^2} + {q^2}}}{{\left( {{p^2} + {q^2}} \right)}} = - 2$
Hence, $\dfrac{{\left( {\dfrac{x}{p} + \dfrac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = - 2$

Therefore, the correct option is D.

Note: The above question is easy to solve with some basic calculations. Just remember that ${i^2} = - 1$ and ${i^3} = - i$ . Also, while comparing the real and imaginary parts, it is important to note that $z = x - iy$, hence, minus sign will be involved. Otherwise, it will lead to miscalculations.