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Complex Numbers question

2025 · 7 Apr · Shift 2 · Q28
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  5. /2025 · 7 Apr · Shift 2 · Q28

Complex Numbers question

2025 · 7 Apr · Shift 2 · Q28

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If the locus of z ∈ ℂ, such that Re (z−12z+i)+Re(z‾−12z‾−i)=2\left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{\overline{z} - 1}{2\overline{z} - i} \right) = 2(2z+iz−1​)+Re(2z−iz−1​)=2, is a circle of radius r and center (a,b)(a, b)(a,b), then 15abr2\frac{15ab}{r^2}r215ab​ is equal to :
  1. A
    16
  2. B
    24
  3. C
    12
  4. D
    18
View written solutionFree

Correct answer: D

Let z=x+iyz=x+iyz=x+iy, so that z‾=x−iy\overline z=x-iyz=x−iy.

We need to solve

ℜ(z−12z+i)+ℜ(z‾−12z‾−i)=2.\Re\left(\frac{z-1}{2z+i}\right)+\Re\left(\frac{\overline z-1}{2\overline z-i}\right)=2.ℜ(2z+iz−1​)+ℜ(2z−iz−1​)=2.

1. Observe the two terms are conjugates

Let

w=z−12z+i.w=\frac{z-1}{2z+i}.w=2z+iz−1​.

Then

w‾=z‾−12z‾−i.\overline w=\frac{\overline z-1}{2\overline z-i}.w=2z−iz−1​.

Hence

ℜ(z−12z+i)=ℜ(z‾−12z‾−i).\Re\left(\frac{z-1}{2z+i}\right)=\Re\left(\frac{\overline z-1}{2\overline z-i}\right).ℜ(2z+iz−1​)=ℜ(2z−iz−1​).

So the given equation becomes

2ℜ(z−12z+i)=22\Re\left(\frac{z-1}{2z+i}\right)=22ℜ(2z+iz−1​)=2

which gives

ℜ(z−12z+i)=1.\Re\left(\frac{z-1}{2z+i}\right)=1.ℜ(2z+iz−1​)=1.

2. Write in terms of x,yx,yx,y

Substitute z=x+iyz=x+iyz=x+iy:

z−12z+i=(x−1)+iy2x+i(2y+1).\frac{z-1}{2z+i}=\frac{(x-1)+iy}{2x+i(2y+1)}.2z+iz−1​=2x+i(2y+1)(x−1)+iy​.

To find its real part, multiply numerator and denominator by the conjugate of the denominator:

(x−1)+iy2x+i(2y+1)⋅2x−i(2y+1)2x−i(2y+1).\frac{(x-1)+iy}{2x+i(2y+1)}\cdot \frac{2x-i(2y+1)}{2x-i(2y+1)}.2x+i(2y+1)(x−1)+iy​⋅2x−i(2y+1)2x−i(2y+1)​.

The numerator becomes

((x−1)+iy)(2x−i(2y+1)).((x-1)+iy)(2x-i(2y+1)).((x−1)+iy)(2x−i(2y+1)).

Its real part is

2x(x−1)+y(2y+1)=2x2−2x+2y2+y.2x(x-1)+y(2y+1)=2x^2-2x+2y^2+y.2x(x−1)+y(2y+1)=2x2−2x+2y2+y.

The denominator is

(2x)2+(2y+1)2=4x2+4y2+4y+1.(2x)^2+(2y+1)^2=4x^2+4y^2+4y+1.(2x)2+(2y+1)2=4x2+4y2+4y+1.

Therefore,

ℜ(z−12z+i)=2x2−2x+2y2+y4x2+4y2+4y+1.\Re\left(\frac{z-1}{2z+i}\right)=\frac{2x^2-2x+2y^2+y}{4x^2+4y^2+4y+1}.ℜ(2z+iz−1​)=4x2+4y2+4y+12x2−2x+2y2+y​.

Given this equals 111, we get

2x2−2x+2y2+y=4x2+4y2+4y+1.2x^2-2x+2y^2+y=4x^2+4y^2+4y+1.2x2−2x+2y2+y=4x2+4y2+4y+1.

So

2x2+2y2+2x+3y+1=0.2x^2+2y^2+2x+3y+1=0.2x2+2y2+2x+3y+1=0.

3. Convert to standard circle form

Divide by 222:

x2+y2+x+32y+12=0.x^2+y^2+x+\frac{3}{2}y+\frac12=0.x2+y2+x+23​y+21​=0.

Complete squares:

(x+12)2−14+(y+34)2−916+12=0.\left(x+\frac12\right)^2-\frac14+\left(y+\frac34\right)^2-\frac{9}{16}+\frac12=0.(x+21​)2−41​+(y+43​)2−169​+21​=0.

Thus

(x+12)2+(y+34)2=516.\left(x+\frac12\right)^2+\left(y+\frac34\right)^2=\frac{5}{16}.(x+21​)2+(y+43​)2=165​.

So the circle has

  • center (a,b)=(−12,−34)(a,b)=\left(-\frac12,-\frac34\right)(a,b)=(−21​,−43​),
  • radius r=54r=\frac{\sqrt5}{4}r=45​​, hence r2=516r^2=\frac{5}{16}r2=165​.

4. Compute the required value

ab=(−12)(−34)=38.ab=\left(-\frac12\right)\left(-\frac34\right)=\frac38.ab=(−21​)(−43​)=83​.

Therefore,

15abr2=15⋅38516=15⋅38⋅165=15⋅25⋅3=18.\frac{15ab}{r^2}=\frac{15\cdot \frac38}{\frac{5}{16}} =15\cdot \frac38\cdot \frac{16}{5} =15\cdot \frac{2}{5}\cdot 3 =18.r215ab​=165​15⋅83​​=15⋅83​⋅516​=15⋅52​⋅3=18.

Hence the correct option is D.

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