Test Index

Complex Numbers Part 1

Section: Mathematics

© examsnet.com

Question : 94 of 100

Marks: +1, -0

If

z=x-iy

z^{\frac{1}{3}}=p+iq

\frac{\frac{x}{p}+\frac{y}{q}}{p^{2}+q^{2}} \text{ is equal to}

$-2$
$-1$
$2$
$1$

Solution:

Given

z^{\frac{1}{3}}=p+iq

\Rightarrow z=(p+iq)^{3}

=p^{3}+(iq)^{3}+3p(iq)(p+iq)

=p^{3}-iq^{3}+3ip^{2}q-3pq^{2}

=p(p^{2}-3q^{2})-iq(q^{2}-3p^{2})

Given that

z=x-iy

\therefore x-iy=p(p^{2}-3q^{2})-iq(q^{2}-3p^{2})

By comparing both sides we get,

\frac{x}{p}=p^{2}-3q^{2}

and

\frac{y}{q}=q^{2}-3p^{2}

\therefore \frac{\frac{2}{p}+\frac{y}{q}}{p^{2}+q^{2}}

=\frac{p^{2}-3q^{2}+q^{2}-3p^{2}}{p^{2}+q^{2}}

=\frac{-2q^{2}-2p^{2}}{p^{2}+q^{2}}

=\frac{-2(q^{2}+p^{2})}{p^{2}+q^{2}}

=-2

© examsnet.com

Go to Question:

Prev Question

Next Question