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Complex Numbers question

2004 · Shift 0 · Q98
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Complex Numbers question

2004 · Shift 0 · Q98

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z=x−iyz = x - iyz=x−iy and z13=p+iq{z^{{1 \over 3}}} = p + iqz31​=p+iq, then (xp+yq)(p2+q2){{\left( {{x \over p} + {y \over q}} \right)} \over {\left( {{p^2} + {q^2}} \right)}}(p2+q2)(px​+qy​)​ is equal to :
  1. A
    - 2
  2. B
    - 1
  3. C
    2
  4. D
    1
View written solutionFree

Correct answer: A

  1. We are given z=x−iyz=x-iyz=x−iy and z1/3=p+iq.z^{1/3}=p+iq.z1/3=p+iq.

So, z=(p+iq)3.z=(p+iq)^3.z=(p+iq)3.

  1. Expand (p+iq)3(p+iq)^3(p+iq)3:
(p+iq)3=p3+3p2(iq)+3p(iq)2+(iq)3.(p+iq)^3=p^3+3p^2(iq)+3p(iq)^2+(iq)^3.(p+iq)3=p3+3p2(iq)+3p(iq)2+(iq)3.

Now,

(iq)2=−q2,eq(iq)3=−iq3.(iq)^2=-q^2, eq(iq)^3=-iq^3.(iq)2=−q2,eq(iq)3=−iq3.

Therefore,

(p+iq)3=(p3−3pq2)+i(3p2q−q3).(p+iq)^3=(p^3-3pq^2)+i(3p^2q-q^3).(p+iq)3=(p3−3pq2)+i(3p2q−q3).

  1. But also z=x−iy.z=x-iy.z=x−iy. Comparing real and imaginary parts: x=p3−3pq2,x=p^3-3pq^2,x=p3−3pq2, and −y=3p2q−q3.-y=3p^2q-q^3.−y=3p2q−q3. Hence, y=q3−3p2q.y=q^3-3p^2q.y=q3−3p2q.

  2. Now compute

(xp+yq)p2+q2.\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{p^2+q^2}.p2+q2(px​+qy​)​.

First,

xp=p3−3pq2p=p2−3q2,\frac{x}{p}=\frac{p^3-3pq^2}{p}=p^2-3q^2,px​=pp3−3pq2​=p2−3q2,

and

yq=q3−3p2qq=q2−3p2.\frac{y}{q}=\frac{q^3-3p^2q}{q}=q^2-3p^2.qy​=qq3−3p2q​=q2−3p2.

So,

xp+yq=(p2−3q2)+(q2−3p2).\frac{x}{p}+\frac{y}{q}=(p^2-3q^2)+(q^2-3p^2).px​+qy​=(p2−3q2)+(q2−3p2).

Simplify:

=−2p2−2q2=−2(p2+q2).= -2p^2-2q^2=-2(p^2+q^2).=−2p2−2q2=−2(p2+q2).
  1. Therefore,
(xp+yq)p2+q2=−2(p2+q2)p2+q2=−2.\frac{\left(\frac{x}{p}+\frac{y}{q}\right)}{p^2+q^2} =\frac{-2(p^2+q^2)}{p^2+q^2}=-2.p2+q2(px​+qy​)​=p2+q2−2(p2+q2)​=−2.
  1. Hence the correct option is −2\boxed{-2}−2​ which is option A.
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