JEE Main 2022 — 26 June, Morning Shift. Previous Year Question.
Problem
Let $A = \left\{z \in \mathbb{C} : \left|\dfrac{z+1}{z-1}\right| < 1\right\}$ and $B = \left\{z \in \mathbb{C} : \arg\left(\dfrac{z-1}{z+1}\right) = \dfrac{2\pi}{3}\right\}$. Then $A \cap B$ is:
Key insight. Both conditions describe geometric loci once $z=x+iy$ is substituted — the modulus condition collapses into a simple half-plane, and the argument condition traces out a full circle (though only part of that circle actually satisfies the original equation, since argument, unlike modulus, is sensitive to direction). The final answer is whatever’s left when both regions overlap.
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Approach
Substitute $z=x+iy$ into the modulus inequality and simplify — this reduces to a straightforward condition on $x$ alone. Then substitute into the argument equation, rationalise the resulting fraction, and separate real and imaginary parts to get the equation of a circle. Finally, check which arc of that circle actually satisfies the original argument value (not just the circle equation), and intersect that arc with the half-plane from the first condition.
Solution
Step 1 — Simplify the modulus condition
$$\left|\frac{z+1}{z-1}\right| < 1 \iff |z+1| < |z-1|$$
Squaring both sides with $z=x+iy$:
$$(x+1)^2+y^2 < (x-1)^2+y^2$$
Expanding, the $x^2$ and $y^2$ terms cancel on both sides, leaving:
$$4x < 0 \implies x < 0$$
So $A$ is simply the open left half-plane: $\text{Re}(z) < 0$.
Step 2 — Set up the argument condition
$$\frac{z-1}{z+1} = \frac{(x-1)+iy}{(x+1)+iy}$$
Multiplying numerator and denominator by the conjugate $(x+1)-iy$:
$$\frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy} = \frac{(x^2+y^2-1) + 2iy}{(x+1)^2+y^2}$$
Step 3 — Extract the argument as an arctangent
Since the denominator is real and positive, the argument of the whole expression is the argument of the numerator:
$$\arg\left(\frac{z-1}{z+1}\right) = \tan^{-1}\left(\frac{2y}{x^2+y^2-1}\right) = \frac{2\pi}{3}$$
Step 4 — Solve for the circle equation
Since $\dfrac{2\pi}{3}$ lies in the second quadrant (of the argument, not the $xy$-plane), the ratio itself must have negative real part and positive imaginary part — i.e. $y > 0$ and $x^2+y^2-1 < 0$. Using $\tan\dfrac{2\pi}{3} = -\sqrt3$:
$$\frac{2y}{x^2+y^2-1} = -\sqrt3 \implies \sqrt3(x^2+y^2-1) = -2y$$
$$x^2+y^2+\frac{2y}{\sqrt3}-1 = 0 \implies x^2+\left(y+\frac{1}{\sqrt3}\right)^2 = \frac{4}{3}$$
This is a circle centred at $\left(0,-\dfrac{1}{\sqrt3}\right)$ with radius $\dfrac{2}{\sqrt3}$ — passing through $z=1$ and $z=-1$, as it must, since the argument expression is undefined exactly there.
Step 5 — Identify which arc actually satisfies the equation
The full circle satisfies the circle equation, but only one arc satisfies $\arg = +\dfrac{2\pi}{3}$ specifically (the other arc gives $\arg = -\dfrac{2\pi}{3}$, i.e. $\dfrac{4\pi}{3}$). Testing the topmost point of this circle, $z = \dfrac{i}{\sqrt3}$, directly in the original expression confirms $\arg\left(\dfrac{z-1}{z+1}\right) = \dfrac{2\pi}{3}$ there — and by the inscribed-angle property, the entire upper arc from $z=-1$ through this point to $z=1$ shares that same argument. This upper arc lies entirely in $y \geq 0$, spanning from the second quadrant (near $z=-1$) through to the first quadrant (near $z=1$).
Step 6 — Intersect with the half-plane $x<0$
Set $A$ restricts to $x<0$, which keeps only the left half of this upper arc — the portion running from $z=-1$ up to the topmost point $\left(0,\frac{1}{\sqrt3}\right)$. Since this entire retained portion has $x<0$ and $y \geq 0$, it lies only in the second quadrant.
Answer
A portion of a circle centred at $\left(0,-\dfrac{1}{\sqrt3}\right)$, radius $\dfrac{2}{\sqrt3}$, lying in the second quadrant only.
Common mistakes
- Assuming the entire circle satisfies the argument condition. The circle equation is necessary but not sufficient — only the arc where the sign of the argument matches $+\dfrac{2\pi}{3}$ (rather than $-\dfrac{2\pi}{3}$) is valid, which is why a direct point-check on the arc is essential.
- Forgetting the arc spans two quadrants before intersecting with $A$. The full valid arc for $B$ actually touches both the first and second quadrants; only after intersecting with $x<0$ does it collapse to the second quadrant alone — jumping straight to “quadrant 2 only” without this check risks the wrong justification even if the final answer is right.
Practise next
- Let $A = \{z : |z-1| < |z+1|\}$ and $B = \{z : \arg\left(\frac{z+1}{z-1}\right) = \frac{\pi}{3}\}$. Find $A \cap B$ using the same substitution-and-arc method.
Show answer
The arc of $x^2+\left(y+\tfrac{1}{\sqrt3}\right)^2=\tfrac43$ lying in the fourth quadrant, from $(0,-\sqrt3)$ to $(1,0)$, both endpoints excluded.
$|z-1|<|z+1|$ is just $\operatorname{Re}(z)>0$ — the open right half-plane.
For $B$, rationalising gives $\dfrac{z+1}{z-1}=\dfrac{(x^2+y^2-1)-2iy}{|z-1|^{2}}$. Setting its argument to $\tfrac{\pi}{3}$ needs both parts positive and $\dfrac{-2y}{x^2+y^2-1}=\sqrt3$, which rearranges to $x^2+y^2+\tfrac{2}{\sqrt3}y-1=0$ with $y<0$ — the circle above, of radius $\tfrac{2}{\sqrt3}$ centred at $\left(0,-\tfrac1{\sqrt3}\right)$, passing through $(\pm1,0)$.
Intersecting with $\operatorname{Re}(z)>0$ keeps only the quarter of that arc from its lowest point $(0,-\sqrt3)$ round to $(1,0)$, where $z=1$ is excluded because it makes the denominator vanish.
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