Let z₁, z₂ and z₃ Be Three Complex Numbers on the Circle |z| = 1 With arg(z₁) = −π/4, arg(z₂) = 0 and arg(z₃) = π/4. If |z₁z̄₂ + z₂z̄₃ + z₃z̄₁|² = α + β√2, α, β ∈ Z, Then α² + β² Is

JEE Main 2025 — 22 January, Morning Shift. Previous Year Question.

Problem

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg(z_1)=-\dfrac{\pi}4$, $\arg(z_2)=0$ and $\arg(z_3)=\dfrac{\pi}4$. If $|z_1\bar z_2+z_2\bar z_3+z_3\bar z_1|^2=\alpha+\beta\sqrt2$, $\alpha,\beta\in\mathbb Z$, then the value of $\alpha^2+\beta^2$ is:

Key insight. With $|z|=1$ for all three points, each is entirely determined by its argument alone — write them directly as $\cos\theta+i\sin\theta$ for the given angles, and the whole problem collapses to arithmetic on three explicit complex numbers.

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Approach

Write $z_1,z_2,z_3$ explicitly using their given arguments on the unit circle. Compute each conjugate, then each of the three products $z_1\bar z_2$, $z_2\bar z_3$, $z_3\bar z_1$, and add them together. Take the modulus squared of the sum, match it against the form $\alpha+\beta\sqrt2$, and compute $\alpha^2+\beta^2$.

Solution

Step 1 — Write z1, z2, z3 explicitly

$$z_1 = \cos\left(-\frac{\pi}4\right)+i\sin\left(-\frac{\pi}4\right) = \frac{1}{\sqrt2}-\frac{i}{\sqrt2}$$

$$z_2 = \cos0+i\sin0 = 1$$

$$z_3 = \cos\frac{\pi}4+i\sin\frac{\pi}4 = \frac{1}{\sqrt2}+\frac{i}{\sqrt2}$$

Step 2 — Compute z1z̄2

Since $\bar z_2=1$:

$$z_1\bar z_2 = z_1 = \frac{1}{\sqrt2}-\frac{i}{\sqrt2}$$

Step 3 — Compute z2z̄3

$$\bar z_3 = \frac{1}{\sqrt2}-\frac{i}{\sqrt2}, \qquad z_2\bar z_3 = 1\cdot\bar z_3 = \frac{1}{\sqrt2}-\frac{i}{\sqrt2}$$

Step 4 — Compute z3z̄1

$$\bar z_1 = \frac{1}{\sqrt2}+\frac{i}{\sqrt2}$$

$$z_3\bar z_1 = \left(\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\right)\left(\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\right) = \left(\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\right)^2 = \frac12+\frac{2i}{2}-\frac12 = i$$

(using $(x+iy)^2=x^2-y^2+2ixy$ with $x=y=\tfrac1{\sqrt2}$, so $x^2-y^2=0$ and $2xy=1$)

Step 5 — Add all three terms

$$z_1\bar z_2+z_2\bar z_3+z_3\bar z_1 = \left(\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right)+\left(\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\right)+i$$

$$= \frac{2}{\sqrt2}-\frac{2i}{\sqrt2}+i = \sqrt2 – i\sqrt2+i = \sqrt2+i(1-\sqrt2)$$

Step 6 — Take the modulus squared

$$\left|\sqrt2+i(1-\sqrt2)\right|^2 = (\sqrt2)^2+(1-\sqrt2)^2 = 2+(1-2\sqrt2+2) = 2+3-2\sqrt2 = 5-2\sqrt2$$

Step 7 — Match against α + β√2

$$\alpha+\beta\sqrt2 = 5-2\sqrt2 \implies \alpha=5,\ \beta=-2$$

Step 8 — Compute α² + β²

$$\alpha^2+\beta^2 = 25+4 = 29$$

Answer

Common mistakes

• Mixing up $z\bar w$ with $\bar z w$. These are generally different complex numbers (conjugates of each other, in fact) — the problem specifically asks for $z_1\bar z_2$ (not $\bar z_1 z_2$), and swapping the conjugate onto the wrong factor changes the sign of the imaginary part in each term.
• Forgetting that $\alpha,\beta\in\mathbb Z$ pins down a unique matching. Since $\sqrt2$ is irrational, an expression like $5-2\sqrt2$ can only be written as $\alpha+\beta\sqrt2$ with integer $\alpha,\beta$ in exactly one way — there’s no ambiguity once the modulus-squared is simplified to this form.

Practise next

• Let $w_1,w_2,w_3$ be complex numbers on $|z|=1$ with $\arg(w_1)=0$, $\arg(w_2)=\dfrac{\pi}3$, $\arg(w_3)=\dfrac{2\pi}3$. If $|w_1\bar w_2+w_2\bar w_3+w_3\bar w_1|^2=p+q\sqrt3$, find $p^2+q^2$, using the same explicit-coordinates approach.