If z=x-iy and z^1/3=p+iq, then ( xp+ yq)/(p^2+q^2) is equal to

Question

If $z=x-iy$ and $z^{1/3}=p+iq$, then $\left(\dfrac xp+\dfrac yq\right)/(p^2+q^2)$ is equal to

Accepted Answer

Given $${z^{{1 \over 3}}} = p + iq$$

$$ \Rightarrow $$ z = (p + iq)3

= p3 + (iq)3 +3p(iq)(p + iq)

= p3 - iq3 +3ip2q - 3pq2

= p(p2 - 3q2) - iq(q2 - 3p2)

Given that $z = x - iy$

$$\therefore$$ $x - iy$ = p(p2 - 3q2) - iq(q2 - 3p2)

By comparing both sides we get,

$${x \over p} = {p^2} - 3{q^2}$$ and $${y \over q} = {q^2} - 3{p^2}$$

$$\therefore$$ $${{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}$$

= $${{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}$$

= $${{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}$$

= $-2$

Solutions · 1