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Complex Numbers question

2021 · 31 Aug · Shift 2 · Q31
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Complex Numbers question

2021 · 31 Aug · Shift 2 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If z is a complex number such that z−iz−1{{z - i} \over {z - 1}}z−1z−i​ is purely imaginary, then the minimum value of | z −-− (3 + 3i) | is :
  1. A
    22−12\sqrt 2 - 122​−1
  2. B
    323\sqrt 232​
  3. C
    626\sqrt 262​
  4. D
    222\sqrt 222​
View written solutionFree

Correct answer: D

  1. Let z=x+iyz=x+iyz=x+iy where x,y∈Rx,y\in\mathbb Rx,y∈R.

We are given that z−iz−1\frac{z-i}{z-1}z−1z−i​ is purely imaginary.

  1. Compute the expression in terms of x,yx,yx,y.

Since z−i=x+i(y−1),z−1=(x−1)+iy,z-i=x+i(y-1),\qquad z-1=(x-1)+iy,z−i=x+i(y−1),z−1=(x−1)+iy, we have z−iz−1=x+i(y−1)(x−1)+iy.\frac{z-i}{z-1}=\frac{x+i(y-1)}{(x-1)+iy}.z−1z−i​=(x−1)+iyx+i(y−1)​.

Multiply numerator and denominator by the conjugate of the denominator: x+i(y−1)(x−1)+iy⋅(x−1)−iy(x−1)−iy.\frac{x+i(y-1)}{(x-1)+iy}\cdot \frac{(x-1)-iy}{(x-1)-iy}.(x−1)+iyx+i(y−1)​⋅(x−1)−iy(x−1)−iy​.

Now expand the numerator: [x+i(y−1)] [(x−1)−iy].[x+i(y-1)]\,[(x-1)-iy].[x+i(y−1)][(x−1)−iy].

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

Imaginary part: −xy+(x−1)(y−1)=−xy+xy−x−y+1=1−x−y.-xy+(x-1)(y-1)= -xy+xy-x-y+1=1-x-y.−xy+(x−1)(y−1)=−xy+xy−x−y+1=1−x−y.

So, z−iz−1=(x2+y2−x−y)+i(1−x−y)(x−1)2+y2.\frac{z-i}{z-1}=\frac{(x^2+y^2-x-y)+i(1-x-y)}{(x-1)^2+y^2}.z−1z−i​=(x−1)2+y2(x2+y2−x−y)+i(1−x−y)​.

  1. For this to be purely imaginary, its real part must be zero. Thus, x2+y2−x−y=0.x^2+y^2-x-y=0.x2+y2−x−y=0.

Complete squares: x2−x+y2−y=0x^2-x+y^2-y=0x2−x+y2−y=0 (x−12)2−14+(y−12)2−14=0\left(x-\frac12\right)^2-\frac14+\left(y-\frac12\right)^2-\frac14=0(x−21​)2−41​+(y−21​)2−41​=0 (x−12)2+(y−12)2=12.\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12.(x−21​)2+(y−21​)2=21​.

This represents a circle with center (12,12)\left(\frac12,\frac12\right)(21​,21​) and radius r=12=12.r=\sqrt{\frac12}=\frac1{\sqrt2}.r=21​​=2​1​.

  1. We need the minimum value of ∣z−(3+3i)∣,|z-(3+3i)|,∣z−(3+3i)∣, which is the distance from the point (3,3)(3,3)(3,3) to this circle.

Distance from (3,3)(3,3)(3,3) to the center (12,12)\left(\frac12,\frac12\right)(21​,21​) is

=\sqrt{\left(\frac52\right)^2+\left(\frac52\right)^2} =\sqrt{\frac{25}{4}+\frac{25}{4}} =\sqrt{\frac{25}{2}} =\frac{5}{\sqrt2}.$$ Hence minimum distance to the circle is $$d-r=\frac{5}{\sqrt2}-\frac1{\sqrt2}=\frac{4}{\sqrt2}=2\sqrt2.$$ So, $$\min |z-(3+3i)|=2\sqrt2.$$ --- 5. Checking options: - A: $2\sqrt2-1$ ❌ - B: $3\sqrt2$ ❌ - C: $6\sqrt2$ ❌ - D: $2\sqrt2$ ✅ Therefore, the correct option is **D**.
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