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)

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.

N neha

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