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Complex Numbers question

2020 · 7 Jan · Shift 1 · Q25
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Complex Numbers question

2020 · 7 Jan · Shift 1 · Q25

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If Reolimits(z−12z+i)=1{\mathop{\rm Re} olimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1Reolimits(2z+iz−1​)=1, where z = x + iy, then the point (x, y) lies on a :
  1. A
    straight line whose slope is 32{3 \over 2}23​
  2. B
    straight line whose slope is −23-{2 \over 3}−32​
  3. C
    circle whose diameter is 52{{\sqrt 5 } \over 2}25​​
  4. D
    circle whose centre is at (−12,−32)\left( { - {1 \over 2}, - {3 \over 2}} \right)(−21​,−23​)
View written solutionFree

Correct answer: C

  1. Let z=x+iy.z=x+iy.z=x+iy. Then z−1=(x−1)+iy,z-1=(x-1)+iy,z−1=(x−1)+iy, and 2z+i=2x+i(2y+1)=2x+i(2y+1).2z+i=2x+i(2y+1)=2x+i(2y+1).2z+i=2x+i(2y+1)=2x+i(2y+1).

  2. We need Re⁡(z−12z+i)=1.\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1.Re(2z+iz−1​)=1. Write the fraction in standard form by multiplying 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)​.

So,

\frac{\big((x-1)+iy\big)\big(2x-i(2y+1)\big)}{(2x)^2+(2y+1)^2}.$$ 3. Extract the real part of the numerator. Using $$(a+ib)(c-id)=(ac+bd)+i(bc-ad),$$ with $$a=x-1,\quad b=y,\quad c=2x,\quad d=2y+1,$$ the real part of the numerator is $$(x-1)(2x)+y(2y+1)=2x^2-2x+2y^2+y.$$ Hence, $$\operatorname{Re}\left(\frac{z-1}{2z+i}\right)= \frac{2x^2-2x+2y^2+y}{4x^2+(2y+1)^2}.$$ Given this equals $1$, we get $$2x^2-2x+2y^2+y=4x^2+(2y+1)^2.$$ 4. Expand the right-hand side: $$(2y+1)^2=4y^2+4y+1,$$ so $$4x^2+(2y+1)^2=4x^2+4y^2+4y+1.$$ Thus, $$2x^2-2x+2y^2+y=4x^2+4y^2+4y+1.$$ Bring all terms to one side: $$0=2x^2+2y^2+2x+3y+1,$$ or $$2x^2+2y^2+2x+3y+1=0.$$ 5. Divide by $2$: $$x^2+y^2+x+\frac{3y}{2}+\frac12=0.$$ Complete squares: $$x^2+x=\left(x+\frac12\right)^2-\frac14,$$ $$y^2+\frac{3}{2}y=\left(y+\frac34\right)^2-\frac{9}{16}.$$ So, $$\left(x+\frac12\right)^2-\frac14+\left(y+\frac34\right)^2-\frac{9}{16}+\frac12=0.$$ Combine constants: $$-\frac14-\frac{9}{16}+\frac12=-\frac{4}{16}-\frac{9}{16}+\frac{8}{16}=-\frac{5}{16}.$$ Hence, $$\left(x+\frac12\right)^2+\left(y+\frac34\right)^2=\frac{5}{16}.$$ 6. This is a circle with - centre $$\left(-\frac12,-\frac34\right),$$ - radius $$\frac{\sqrt5}{4}.$$ Therefore its diameter is $$2\cdot \frac{\sqrt5}{4}=\frac{\sqrt5}{2}.$$ 7. Check options: - A: straight line, slope $\frac32$ ❌ - B: straight line, slope $-\frac23$ ❌ - C: circle whose diameter is $\frac{\sqrt5}{2}$ ✅ - D: centre at $\left(-\frac12,-\frac32\right)$ ❌ (actual centre is $\left(-\frac12,-\frac34\right)$) Therefore, the correct option is **C**.
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