Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2024 · 6 Apr · Shift 2 · Q31
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2024 · 6 Apr · Shift 2 · Q31

Complex Numbers question

2024 · 6 Apr · Shift 2 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z1,z2z_1, z_2z1​,z2​ are two distinct complex number such that ∣z1−2z212−z1zˉ2∣=2\left|\frac{z_1-2 z_2}{\frac{1}{2}-z_1 \bar{z}_2}\right|=2​21​−z1​zˉ2​z1​−2z2​​​=2, then
  1. A
    either z1z_1z1​ lies on a circle of radius 12\frac{1}{2}21​ or z2z_2z2​ lies on a circle of radius 1.
  2. B
    z1z_1z1​ lies on a circle of radius 12\frac{1}{2}21​ and z2z_2z2​ lies on a circle of radius 1.
  3. C
    either z1z_1z1​ lies on a circle of radius 1 or z2z_2z2​ lies on a circle of radius 12\frac{1}{2}21​.
  4. D
    both z1z_1z1​ and z2z_2z2​ lie on the same circle.
View written solutionFree

Correct answer: C

  1. We are given
∣z1−2z212−z1z2‾∣=2.\left|\frac{z_1-2z_2}{\frac12-z_1\overline{z_2}}\right|=2.​21​−z1​z2​​z1​−2z2​​​=2.

Since modulus of a quotient equals quotient of moduli,

∣z1−2z2∣=2∣12−z1z2‾∣.|z_1-2z_2|=2\left|\frac12-z_1\overline{z_2}\right|.∣z1​−2z2​∣=2​21​−z1​z2​​​.

Now note that

2∣12−z1z2‾∣=∣1−2z1z2‾∣.2\left|\frac12-z_1\overline{z_2}\right|=|1-2z_1\overline{z_2}|.2​21​−z1​z2​​​=∣1−2z1​z2​​∣.

So the condition becomes

∣z1−2z2∣=∣1−2z1z2‾∣.|z_1-2z_2|=|1-2z_1\overline{z_2}|.∣z1​−2z2​∣=∣1−2z1​z2​​∣.
  1. Square both sides:
∣z1−2z2∣2=∣1−2z1z2‾∣2.|z_1-2z_2|^2=|1-2z_1\overline{z_2}|^2.∣z1​−2z2​∣2=∣1−2z1​z2​​∣2.

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,

∣1−2z1z2‾∣2=(1−2z1z2‾)(1−2z1‾z2)|1-2z_1\overline{z_2}|^2=(1-2z_1\overline{z_2})(1-2\overline{z_1}z_2)∣1−2z1​z2​​∣2=(1−2z1​z2​​)(1−2z1​​z2​) =1−2z1z2‾−2z1‾z2+4∣z1∣2∣z2∣2.=1-2z_1\overline{z_2}-2\overline{z_1}z_2+4|z_1|^2|z_2|^2.=1−2z1​z2​​−2z1​​z2​+4∣z1​∣2∣z2​∣2.
  1. Equate the two expressions:
∣z1∣2−2z1z2‾−2z1‾z2+4∣z2∣2=1−2z1z2‾−2z1‾z2+4∣z1∣2∣z2∣2.|z_1|^2-2z_1\overline{z_2}-2\overline{z_1}z_2+4|z_2|^2 =1-2z_1\overline{z_2}-2\overline{z_1}z_2+4|z_1|^2|z_2|^2.∣z1​∣2−2z1​z2​​−2z1​​z2​+4∣z2​∣2=1−2z1​z2​​−2z1​​z2​+4∣z1​∣2∣z2​∣2.

The common terms cancel, giving

∣z1∣2+4∣z2∣2=1+4∣z1∣2∣z2∣2.|z_1|^2+4|z_2|^2=1+4|z_1|^2|z_2|^2.∣z1​∣2+4∣z2​∣2=1+4∣z1​∣2∣z2​∣2.

Rearrange:

4∣z1∣2∣z2∣2−∣z1∣2−4∣z2∣2+1=0.4|z_1|^2|z_2|^2-|z_1|^2-4|z_2|^2+1=0.4∣z1​∣2∣z2​∣2−∣z1​∣2−4∣z2​∣2+1=0.

Factor:

(∣z1∣2−1)(4∣z2∣2−1)=0.(|z_1|^2-1)(4|z_2|^2-1)=0.(∣z1​∣2−1)(4∣z2​∣2−1)=0.

Hence,

∣z1∣2=1or4∣z2∣2=1.|z_1|^2=1 \quad \text{or} \quad 4|z_2|^2=1.∣z1​∣2=1or4∣z2​∣2=1.

So,

∣z1∣=1or∣z2∣=12.|z_1|=1 \quad \text{or} \quad |z_2|=\frac12.∣z1​∣=1or∣z2​∣=21​.
  1. Interpret geometrically:
  • ∣z1∣=1|z_1|=1∣z1​∣=1 means z1z_1z1​ lies on the circle centered at origin with radius 111.
  • ∣z2∣=12|z_2|=\frac12∣z2​∣=21​ means z2z_2z2​ lies on the circle centered at origin with radius 12\frac1221​.

Therefore the correct statement is:

either z1 lies on a circle of radius 1 or z2 lies on a circle of radius 12.\text{either } z_1 \text{ lies on a circle of radius } 1 \text{ or } z_2 \text{ lies on a circle of radius } \frac12.either z1​ lies on a circle of radius 1 or z2​ lies on a circle of radius 21​.
  1. Check options:
  • A: false
  • B: false
  • C: true
  • D: false

So the correct option is C.

PreviousNext

More from Complex Numbers

  • Let z be a complex number such that ∣z+2∣=1 and lm(z+2z+1​)=51​. Then the value of ∣Re(z+2​)∣ is2024 · MCQ
  • If the set R={(a,b):a+5b=42,a,b∈N} has m elements and ∑n=1m​(1−in!)=x+iy, where i=−1​, then the value of m+x+y is2024 · MCQ
  • The sum of all possible values of θ∈[−π,2π], for which 1−2icosθ1+icosθ​ is purely imaginary, is equal to :2024 · MCQ
  • The sum of the square of the modulus of the elements in the set {z=a+ib:a,b∈Z,z∈C,∣z−1∣≤1,∣z−5∣≤∣z−5i∣} is ​.2024 · Numerical
  • Let z be a complex number such that the real part of z+2iz−2i​ is zero. Then, the maximum value of ∣z−(6+8i)∣ is equal to2024 · MCQ
  • If S={z∈C:∣z−i∣=∣z+i∣=∣z−1∣}, then, n(S) is :2024 · MCQ
  • If α satisfies the equation x2+x+1=0 and (1+α)7=A+Bα+Cα2,A,B,C⩾0, then 5(3A−2B−C) is equal to ​.2024 · Numerical
  • Let the complex numbers α and αˉ1​ lie on the circles ∣z−z0​∣2=4 and ∣z−z0​∣2=16 respectively, where z0​=1+i. Then, the value of 100∣α∣2 is ​.2024 · Numerical