JEE Main 2025 — 23 January, Evening Shift. Previous Year Question.
Problem
The number of complex numbers $z$, satisfying $|z|=1$ and $\left|\dfrac{z}{\bar z}+\dfrac{\bar z}{z}\right|=1$, is:
Key insight. Because $|z|=1$ means $z\bar z=1$, the fraction $\dfrac{z}{\bar z}$ simplifies to $z^2$ outright — no need to carry $\bar z$ through the rest of the algebra at all. The whole problem collapses to a simple system in $x^2$ and $y^2$.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Write $z=x+iy$ with $x^2+y^2=1$. Since $z\bar z=|z|^2=1$, rewrite $\dfrac{z}{\bar z}=\dfrac{z\cdot z}{\bar z\cdot z}=\dfrac{z^2}{1}=z^2$, and similarly $\dfrac{\bar z}{z}=\bar z^2$. Add these and take the modulus to get an equation purely in $x^2-y^2$, then solve it alongside $x^2+y^2=1$.
Solution
Step 1 — Simplify z/z̄ and z̄/z using |z|=1
Since $|z|=1$, $\bar z=\dfrac{1}{z}$, so:
$$\frac{z}{\bar z}=\frac{z}{1/z}=z^2, \qquad \frac{\bar z}{z}=\frac{1/z}{z}=\bar z^2$$
Step 2 — Add and expand with z = x + iy
$$z^2+\bar z^2=(x+iy)^2+(x-iy)^2=2(x^2-y^2)$$
(the cross terms $2ixy$ and $-2ixy$ cancel)
Step 3 — Apply the given modulus condition
$$\left|2(x^2-y^2)\right|=1 \implies x^2-y^2=\pm\frac{1}{2}$$
Step 4 — Solve alongside x² + y² = 1: Case x² − y² = 1/2
$$x^2+y^2=1, \quad x^2-y^2=\frac{1}{2}$$
Adding: $2x^2=\dfrac{3}{2} \implies x^2=\dfrac{3}{4}$. Subtracting: $2y^2=\dfrac{1}{2}\implies y^2=\dfrac{1}{4}$.
So $x=\pm\dfrac{\sqrt3}{2}$, $y=\pm\dfrac12$ — every sign combination works, giving $4$ points.
Step 5 — Case x² − y² = −1/2
$$x^2+y^2=1, \quad x^2-y^2=-\frac{1}{2}$$
Adding: $2x^2=\dfrac12 \implies x^2=\dfrac14$. Subtracting: $2y^2=\dfrac32 \implies y^2=\dfrac34$.
So $x=\pm\dfrac12$, $y=\pm\dfrac{\sqrt3}{2}$ — again every sign combination works, giving another $4$ points.
Step 6 — Total count
$$4+4=8$$
Answer
$$8$$
Common mistakes
- Working with $\bar z$ directly instead of eliminating it via $|z|=1$. Substituting $\bar z=x-iy$ everywhere and clearing fractions the long way works but invites algebra slips — using $\bar z=1/z$ (valid only because $|z|=1$) shortens the whole problem to a few lines.
- Forgetting that each case yields 4 points, not 1. Once $x^2$ and $y^2$ are pinned down, both $x$ and $y$ independently take a $\pm$ sign, so each case contributes $2\times2=4$ solutions, not just one representative point.
Practise next
- Find the number of complex numbers $z$ satisfying $|z|=2$ and $\left|\dfrac{z}{\bar z}-\dfrac{\bar z}{z}\right|=1$, using the same $z\bar z=|z|^2$ simplification.
Show answer
$8$. On $|z|=2$ write $z=2e^{i\theta}$. Then $\dfrac{z}{\bar z}=e^{2i\theta}$ and $\dfrac{\bar z}{z}=e^{-2i\theta}$.
Their difference is $2i\sin2\theta$, so the condition is $2\left|\sin2\theta\right|=1$, that is $\left|\sin2\theta\right|=\tfrac12$.
As $\theta$ runs over $[0,2\pi)$, $2\theta$ covers two full turns, and $\left|\sin\phi\right|=\tfrac12$ has four solutions per turn. So there are $8$ values of $\theta$, and $8$ complex numbers.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.