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If $z= x-iy\: and \: z^{13}= p+iq,$ then $\frac{\left ( \frac{x}{p} +\frac{y}{q}\right )}{\left ( p^{2}+q^{2} \right )}$ is equal to

Option 1)
$2$

Option 2)
$-1$

Option 3)
$1$

Option 4)
$-2$

Answers (2)

best_answer

As we learnt

Equality in Complex Numbers -

z=x+iy & w=a+ib are equal iff x=a & y=b

- wherein

Two complex numbers are equal iff real parts as well as imaginary parts are equal.

$z=x-iy$ and $z^{1/3}=p+iq$

$\therefore\ \; z=(z+iq)^{3}$

$\therefore\ \; x-iy=p^{3}+(iq)^{3}+3p.iq(p+iq)=p^{3}+5^{3}q^{3}+ip^{2}q-3pq^{2}$

$x-iy=(p^{3}-3pq^{2})+i(3p^{2}q-q^{3})$

Compare: $p^{3}-3pq^{2}=x$

$p(p^{2}-3q^{2})=x$

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

and $q(3p^{2}-q^{2})=-y$

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

$\therefore\ \; \frac{\frac{x}{p}+\frac{y}{q}}{p^{2}+q^{2}}=\frac{p^{2}-3q^{2}+q^{2}-3p^{2}}{(p^{2}+q^{2})}=\frac{-2(p^{2}+q^{2})}{p^{2}+q^{2}}=-2$

Correct option is 4.

Option 1)

$2$

This is an incorrect option.

Option 2)

$-1$

This is an incorrect option.

Option 3)

$1$

This is an incorrect option.

Option 4)

$-2$

This is the correct option.

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