Verified Solution • 2025–26 Session moderate

QJEEM04Ma72d8: JEE Main 2004 Mathematics - Class 11 JEEM04Ma72d8 Solution: If and , then is equal to :

If z=x−iyz = x - iy and z13=p+iq{z^{{1 \over 3}}} = p + iq , then

(xp+yq)(p2+q2){{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}} is equal to :

Multiple Choice Options

  • A -2
  • B -1
  • C 2
  • D 1

Step-by-Step Verified Solution

Given z13=p+iq{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−iyz = x - iy

∴\therefore x−iyx - iy = p(p2 - 3q2) - iq(q2 - 3p2)

By comparing both sides we get,

xp=p2−3q2{x \over p} = {p^2} - 3{q^2} and yq=q2−3p2{y \over q} = {q^2} - 3{p^2}

∴\therefore (xp+yq)(p2+q2){{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}

= p2−3q2+q2−3p2p2+q2{{{p^2} - 3{q^2} + {q^2} - 3{p^2}} \over {{p^2} + {q^2}}}

= −2q2−2p2p2+q2{{ - 2{q^2} - 2{p^2}} \over {{p^2} + {q^2}}}

= −2(q2+p2)p2+q2{{ - 2\left( {{q^2} + {p^2}} \right)} \over {{p^2} + {q^2}}}

= −2-2

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