Let A = \ z \in C:| z + 1 \over z - 1 | \lt 1 \ and B = \ z \in C:\arg ( z - 1 \over z + 1 ) = 2\pi \over 3 \ . Then A \cap B is :

a portion of a circle centred at ( 0, - 1 \over \sqrt 3 ) that lies in the second and third quadrants only

a portion of a circle centred at ( 0, - 1 \over \sqrt 3 ) that lies in the second quadrant only

a portion of a circle of radius 2 \over \sqrt 3 that lies in the third quadrant only

Complex Numbers PYQ — JEE Main 2022 · 26 Jun · Shift 1 · Q22 with Solution

View written solutionFree

Correct answer: A

Find the region $A$

We have

A=\left\{z\in \mathbb C:\left|\frac{z+1}{z-1}\right|<1\right\}.

Let $z=x+iy$ . Then

\left|\frac{z+1}{z-1}\right|<1 \iff |z+1|<|z-1|.

This means the point $z$ is closer to $-1$ than to $1$ on the real axis.

Now,

|z+1|^2<(z-1|)^2

actually means

|z+1|^2<|z-1|^2.

So,

(x+1)^2+y^2<(x-1)^2+y^2.

Cancelling $y^2$ ,

(x+1)^2<(x-1)^2

x^2+2x+1<x^2-2x+1

4x<0

x<0.

Hence,

A=\{z=x+iy:x<0\},

which is the left half-plane.

Find the locus $B$

Given

B=\left\{z\in \mathbb C:\arg\left(\frac{z-1}{z+1}\right)=\frac{2\pi}{3}\right\}.

Let

w=\frac{z-1}{z+1}.

Then $\arg w=\frac{2\pi}{3}$ , so $w$ lies on the ray making angle $\frac{2\pi}{3}$ with the positive real axis.

Write

\frac{z-1}{z+1}=re^{i2\pi/3},\qquad r>0.

Then

z-1=re^{i2\pi/3}(z+1).

z\bigl(1-re^{i2\pi/3}\bigr)=1+re^{i2\pi/3},

z=\frac{1+re^{i2\pi/3}}{1-re^{i2\pi/3}}.

To identify the locus, use the standard argument property:

\arg\left(\frac{z-1}{z+1}\right)=\arg(z-1)-\arg(z+1)=\frac{2\pi}{3}.

This means the oriented angle between the segments joining $z$ to $1$ and $-1$ is constant. Therefore, the locus is an arc of a circle passing through $1$ and $-1$ .

Let the circle subtend angle $\frac{2\pi}{3}$ at point $z$ . If the chord is the segment from $-1$ to $1$ , then its length is $2$ .

For a circle of radius $R$ with chord length $2$ subtending inscribed angle $\frac{2\pi}{3}$ ,

2=2R\sin\left(\frac{2\pi}{3}\right)=2R\cdot \frac{\sqrt3}{2}=R\sqrt3.

Hence,

R=\frac{2}{\sqrt3}.

The center lies on the perpendicular bisector of the chord joining $-1$ and $1$ , i.e. on the imaginary axis. Let the center be $(0,k)$ . Since distance to $(1,0)$ equals $R$ ,

1+k^2=\left(\frac{2}{\sqrt3}\right)^2=\frac43.

Thus,

k^2=\frac13 \implies k=\pm \frac{1}{\sqrt3}.

Now determine which one corresponds to argument $\frac{2\pi}{3}$ . Take a test point on the lower arc, say the lowest point of the circle centered at $\left(0,-\frac1{\sqrt3}\right)$ . That point gives the required positive angle $\frac{2\pi}{3}$ . Hence the relevant circle is

x^2+\left(y+\frac1{\sqrt3}\right)^2=\frac43,

with center

\left(0,-\frac1{\sqrt3}\right).

But $B$ is not the whole circle; it is the arc excluding $z=\pm1$ (since the expression is undefined at $z=-1$ and zero at $z=1$ , whose argument is not defined).

This arc lies in the second, third, and fourth quadrants. Indeed, the circle reaches up to

y_{\max}=-\frac1{\sqrt3}+\frac{2}{\sqrt3}=\frac1{\sqrt3}>0,

so part is in quadrant II; it also extends below the x-axis into quadrants III and IV.

Intersect with $A$

Since

A=\{x<0\},

we keep only the part of the above arc lying in the left half-plane.

The circle intersects the $y$ -axis at $x=0$ :

\left(y+\frac1{\sqrt3}\right)^2=\frac43

y+\frac1{\sqrt3}=\pm \frac{2}{\sqrt3}

y=\frac1{\sqrt3},\quad y=-\sqrt3.

Thus, the left-half portion of the relevant arc runs from $(0,\frac1{\sqrt3})$ down to $(0,-\sqrt3)$ through the left side of the circle, i.e. through quadrants II and III.

Therefore,

A\cap B

is a portion of the circle centered at $\left(0,-\frac1{\sqrt3}\right)$ lying in the second and third quadrants.

So the correct option should be:

\boxed{\text{A}}

Compare with stored answer

Stored correct answer: $\text{B}$

Our derived answer: $\text{A}$

They do not match. The reason is that $A$ is the entire left half-plane $x<0$ , so the intersection must include both the second and third quadrant parts of the arc, not only the second quadrant part.

More from Complex Numbers

View written solutionFree

Correct answer: A

Find the region $A$

We have

A=\left\{z\in \mathbb C:\left|\frac{z+1}{z-1}\right|<1\right\}.

Let $z=x+iy$ . Then

\left|\frac{z+1}{z-1}\right|<1 \iff |z+1|<|z-1|.

This means the point $z$ is closer to $-1$ than to $1$ on the real axis.

Now,

|z+1|^2<(z-1|)^2

actually means

|z+1|^2<|z-1|^2.

So,

(x+1)^2+y^2<(x-1)^2+y^2.

Cancelling $y^2$ ,

(x+1)^2<(x-1)^2

x^2+2x+1<x^2-2x+1

4x<0

x<0.

Hence,

A=\{z=x+iy:x<0\},

which is the left half-plane.

Find the locus $B$

Given

B=\left\{z\in \mathbb C:\arg\left(\frac{z-1}{z+1}\right)=\frac{2\pi}{3}\right\}.

Let

w=\frac{z-1}{z+1}.

Then $\arg w=\frac{2\pi}{3}$ , so $w$ lies on the ray making angle $\frac{2\pi}{3}$ with the positive real axis.

Write

\frac{z-1}{z+1}=re^{i2\pi/3},\qquad r>0.

Then

z-1=re^{i2\pi/3}(z+1).

z\bigl(1-re^{i2\pi/3}\bigr)=1+re^{i2\pi/3},

z=\frac{1+re^{i2\pi/3}}{1-re^{i2\pi/3}}.

To identify the locus, use the standard argument property:

\arg\left(\frac{z-1}{z+1}\right)=\arg(z-1)-\arg(z+1)=\frac{2\pi}{3}.

This means the oriented angle between the segments joining $z$ to $1$ and $-1$ is constant. Therefore, the locus is an arc of a circle passing through $1$ and $-1$ .

Let the circle subtend angle $\frac{2\pi}{3}$ at point $z$ . If the chord is the segment from $-1$ to $1$ , then its length is $2$ .

For a circle of radius $R$ with chord length $2$ subtending inscribed angle $\frac{2\pi}{3}$ ,

2=2R\sin\left(\frac{2\pi}{3}\right)=2R\cdot \frac{\sqrt3}{2}=R\sqrt3.

Hence,

R=\frac{2}{\sqrt3}.

The center lies on the perpendicular bisector of the chord joining $-1$ and $1$ , i.e. on the imaginary axis. Let the center be $(0,k)$ . Since distance to $(1,0)$ equals $R$ ,

1+k^2=\left(\frac{2}{\sqrt3}\right)^2=\frac43.

Thus,

k^2=\frac13 \implies k=\pm \frac{1}{\sqrt3}.

x^2+\left(y+\frac1{\sqrt3}\right)^2=\frac43,

with center

\left(0,-\frac1{\sqrt3}\right).

But $B$ is not the whole circle; it is the arc excluding $z=\pm1$ (since the expression is undefined at $z=-1$ and zero at $z=1$ , whose argument is not defined).

This arc lies in the second, third, and fourth quadrants. Indeed, the circle reaches up to

y_{\max}=-\frac1{\sqrt3}+\frac{2}{\sqrt3}=\frac1{\sqrt3}>0,

so part is in quadrant II; it also extends below the x-axis into quadrants III and IV.

Intersect with $A$

Since

A=\{x<0\},

we keep only the part of the above arc lying in the left half-plane.

The circle intersects the $y$ -axis at $x=0$ :

\left(y+\frac1{\sqrt3}\right)^2=\frac43

y+\frac1{\sqrt3}=\pm \frac{2}{\sqrt3}

y=\frac1{\sqrt3},\quad y=-\sqrt3.

Thus, the left-half portion of the relevant arc runs from $(0,\frac1{\sqrt3})$ down to $(0,-\sqrt3)$ through the left side of the circle, i.e. through quadrants II and III.

Therefore,

A\cap B

is a portion of the circle centered at $\left(0,-\frac1{\sqrt3}\right)$ lying in the second and third quadrants.

So the correct option should be:

\boxed{\text{A}}

Compare with stored answer

Stored correct answer: $\text{B}$

Our derived answer: $\text{A}$

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