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Complex Numbers question

2025 · 7 Apr · Shift 1 · Q38
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  5. /2025 · 7 Apr · Shift 1 · Q38

Complex Numbers question

2025 · 7 Apr · Shift 1 · Q38

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Among the statements (S1) : The set {z∈C−{−i}:∣z∣=1\left\{z \in \mathbb{C}-\{-i\}:|z|=1\right.{z∈C−{−i}:∣z∣=1 and z−iz+i\frac{z-i}{z+i}z+iz−i​ is purely real }\}} contains exactly two elements, and (S2) : The set {z∈C−{−1}:∣z∣=1\left\{z \in \mathbb{C}-\{-1\}:|z|=1\right.{z∈C−{−1}:∣z∣=1 and z−1z+1\frac{z-1}{z+1}z+1z−1​ is purely imaginary }\}} contains infinitely many elements.
  1. A
    both are incorrect
  2. B
    both are correct
  3. C
    only (S2) is correct
  4. D
    only (S1) is correct
View written solutionFree

Correct answer: C

  1. Statement (S1)

We need to find

{z∈C∖{−i}:∣z∣=1 and z−iz+i is purely real}.\left\{z\in \mathbb C\setminus\{-i\}: |z|=1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\right\}.{z∈C∖{−i}:∣z∣=1 and z+iz−i​ is purely real}.

Let z=x+iy,x,y∈R,z=x+iy, \qquad x,y\in \mathbb R,z=x+iy,x,y∈R, with x2+y2=1x^2+y^2=1x2+y2=1 because ∣z∣=1|z|=1∣z∣=1.

Now compute

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

Multiply numerator and denominator by the conjugate of the denominator:

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

The numerator becomes

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

Expanding,

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

Now,

−ix(y+1)+ix(y−1)=ix[(y−1)−(y+1)]=−2ix,-ix(y+1)+ix(y-1)=ix[(y-1)-(y+1)]=-2ix,−ix(y+1)+ix(y−1)=ix[(y−1)−(y+1)]=−2ix,

and

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

So numerator is

x2+y2−1−2ix.x^2+y^2-1-2ix.x2+y2−1−2ix.

Since x2+y2=1x^2+y^2=1x2+y2=1, this reduces to

−2ix.-2ix.−2ix.

Denominator is

x2+(y+1)2,x^2+(y+1)^2,x2+(y+1)2,

which is real.

Hence

z−iz+i=−2ixx2+(y+1)2,\frac{z-i}{z+i}=\frac{-2ix}{x^2+(y+1)^2},z+iz−i​=x2+(y+1)2−2ix​,

which is purely imaginary for every such zzz (except where denominator is zero, i.e. z=−iz=-iz=−i, already excluded).

For it to be purely real, its imaginary part must vanish. Since the expression is purely imaginary, this happens only when it is actually 000, i.e. x=0.x=0.x=0. Then from x2+y2=1x^2+y^2=1x2+y2=1, y=±1.y=\pm 1.y=±1. So the possible points are z=i, −i.z=i,\,-i.z=i,−i. But z=−iz=-iz=−i is excluded. Therefore only z=iz=iz=i is allowed.

So the set contains exactly one element, not two.

Therefore, (S1) is incorrect.

  1. Statement (S2)

We need to find

{z∈C∖{−1}:∣z∣=1 and z−1z+1 is purely imaginary}.\left\{z\in \mathbb C\setminus\{-1\}: |z|=1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\right\}.{z∈C∖{−1}:∣z∣=1 and z+1z−1​ is purely imaginary}.

Again let

Then

z−1z+1=(x−1)+iy(x+1)+iy.\frac{z-1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy}.z+1z−1​=(x+1)+iy(x−1)+iy​.

Multiply by the conjugate of the denominator:

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

Numerator:

((x−1)+iy)((x+1)−iy).((x-1)+iy)((x+1)-iy).((x−1)+iy)((x+1)−iy).

Expanding,

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

Now,

−iy(x−1)+iy(x+1)=iy[(x+1)−(x−1)]=2iy,-iy(x-1)+iy(x+1)=iy[(x+1)-(x-1)]=2iy,−iy(x−1)+iy(x+1)=iy[(x+1)−(x−1)]=2iy,

and

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

So numerator becomes

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

Thus

z−1z+1=2iy(x+1)2+y2,\frac{z-1}{z+1}=\frac{2iy}{(x+1)^2+y^2},z+1z−1​=(x+1)2+y22iy​,

which is purely imaginary for every point on the unit circle except where denominator is zero, i.e. z=−1z=-1z=−1, already excluded.

So every point on the unit circle except −1-1−1 satisfies the condition.

That set has infinitely many elements.

Therefore, (S2) is correct.

  1. Conclusion
  • (S1) is incorrect
  • (S2) is correct

Hence the correct option is C\boxed{\text{C}}C​ which says only (S2) is correct.

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