Question
If $$z = x - i y$$ and $${z^{\frac{1}{3}}} = p + iq,{\text{then }}\frac{{\left( {\frac{x}{p} + \frac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}}$$
equal to
A.
$$ - 2$$
B.
$$ - 1$$
C.
$$2$$
D.
$$1$$
Answer :
$$ - 2$$
Solution :
$$\eqalign{
& {z^{\frac{1}{3}}} = p + iq \cr
& \Rightarrow \,\,z = {p^3} + {\left( {iq} \right)^3} + 3p\left( {iq} \right)\left( {p + iq} \right) \cr
& \Rightarrow \,\,x - iy = {p^3} - 3p{q^2} + i\left( {3{p^2}q - {q^3}} \right) \cr
& \therefore \,\,x = {p^3} - 3p{q^2} \cr
& \Rightarrow \,\,\frac{x}{p} = {p^2} - 3{q^2} \cr
& y = {q^3} - 3{p^2}q \cr
& \Rightarrow \,\,\frac{y}{q} = {q^2} - 3{p^2} \cr
& \therefore \,\,\frac{x}{p} + \frac{y}{q} = - 2{p^2}\, - \,2{q^2} \cr
& \therefore \,\,\frac{{\left( {\frac{x}{p} + \frac{y}{q}} \right)}}{{\left( {{p^2} + {q^2}} \right)}} = - 2 \cr} $$