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Complex Numbers question

2015 · Shift 0 · Q41
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Complex Numbers question

2015 · Shift 0 · Q41

JEE MainMathematicsComplex NumbersMCQ+4 / −1
A complex number z is said to be unimodular if  ∣z∣=1\,\left| z \right| = 1∣z∣=1. Suppose z1{z_1}z1​ and z2{z_2}z2​ are complex numbers such that z1−2z22−z1z2‾{{{z_1} - 2{z_2}} \over {2 - {z_1}\overline {{z_2}} }}2−z1​z2​​z1​−2z2​​ is unimodular and z2{z_2}z2​ is not unimodular. Then the point z1{z_1}z1​ lies on a :
  1. A
    circle of radius 2.
  2. B
    circle of radius 2{\sqrt 2 }2​.
  3. C
    straight line parallel to x-axis
  4. D
    straight line parallel to y-axis.
View written solutionFree

Correct answer: A

  1. Let w=z1−2z22−z1z2‾.w=\frac{z_1-2z_2}{2-z_1\overline{z_2}}.w=2−z1​z2​​z1​−2z2​​. Given that www is unimodular, we have ∣w∣=1.|w|=1.∣w∣=1. So, ∣z1−2z2∣=∣2−z1z2‾∣.\left|z_1-2z_2\right|=\left|2-z_1\overline{z_2}\right|.∣z1​−2z2​∣=∣2−z1​z2​​∣.

  2. Square both sides: ∣z1−2z2∣2=∣2−z1z2‾∣2.|z_1-2z_2|^2=|2-z_1\overline{z_2}|^2.∣z1​−2z2​∣2=∣2−z1​z2​​∣2. Now expand both sides.

For the left side, ∣z1−2z2∣2=(z1−2z2)(z1‾−2z2‾)|z_1-2z_2|^2=(z_1-2z_2)(\overline{z_1}-2\overline{z_2})∣z1​−2z2​∣2=(z1​−2z2​)(z1​​−2z2​​) =∣z1∣2−2z1z2‾−2z1‾z2+4∣z2∣2.=|z_1|^2-2z_1\overline{z_2}-2\overline{z_1}z_2+4|z_2|^2.=∣z1​∣2−2z1​z2​​−2z1​​z2​+4∣z2​∣2.

For the right side, ∣2−z1z2‾∣2=(2−z1z2‾)(2−z1‾z2)|2-z_1\overline{z_2}|^2=(2-z_1\overline{z_2})(2-\overline{z_1}z_2)∣2−z1​z2​​∣2=(2−z1​z2​​)(2−z1​​z2​) =4−2z1z2‾−2z1‾z2+∣z1∣2∣z2∣2.=4-2z_1\overline{z_2}-2\overline{z_1}z_2+|z_1|^2|z_2|^2.=4−2z1​z2​​−2z1​​z2​+∣z1​∣2∣z2​∣2.

  1. Equate them:
=4-2z_1\overline{z_2}-2\overline{z_1}z_2+|z_1|^2|z_2|^2.$$ The middle terms cancel, giving $$|z_1|^2+4|z_2|^2=4+|z_1|^2|z_2|^2.$$ Rearrange: $$|z_1|^2-|z_1|^2|z_2|^2=4-4|z_2|^2,$$ $$|z_1|^2(1-|z_2|^2)=4(1-|z_2|^2).$$ 4. Since $z_2$ is not unimodular, $$|z_2|\ne 1 \implies 1-|z_2|^2\ne 0.$$ So we can divide by $1-|z_2|^2$: $$|z_1|^2=4.$$ Hence, $$|z_1|=2.$$ 5. The locus of $z_1$ is therefore the circle centered at the origin with radius $2$. So the correct option is: $$\boxed{\text{A: circle of radius }2}.$$ 6. Comparison with stored answer: Stored correct answer = A. Our derived answer = A. Hence they agree.
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