Let integers $a, b \in [-3,3]$ be such that $a+b \neq 0$. Then the number of all possible ordered pairs $(a, b)$, for which $\left|\frac{z-a}{z+b}\right|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^{2} \\ \omega & z+\omega^{2} & 1 \\ \omega^{2} & 1 & z+\omega\end{array}\right|=1$, $z \in \mathbb{C}$, where $\omega$ and $\omega^{2}$ are the roots of $x^{2}+x+1=0$, is equal to

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Complex Numbers

Let integers

a, b \in [- 3, 3]

be such that

a + b \neq = 0

. Then the number of all possible ordered pairs

(a, b)

, for which

\frac{z - a}{z + b} = 1

and

z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = 1

z \in C

, where

ω

and

ω^{2}

are the roots of

x^{2} + x + 1 = 0

, is equal to

Answer: 10

Expanding the determinant (using

1 + ω + ω^{2} = 0

and

ω^{3} = 1

), row and column reduction shows it equals

z^{3}

z + 1 ω ω^{2} ω z + ω^{2} 1 ω^{2} 1 z + ω = z^{3}

Setting

z^{3} = 1

gives

z \in {1, ω, ω^{2}}

.

The condition

\frac{z - a}{z + b} = 1

means

z

is equidistant from the real points

a

and

- b

, i.e.

Re (z) = \frac{a - b}{2}

.

For

z = 1

Re (z) = 1 \Rightarrow a - b = 2

.
For

z = ω

ω^{2}

Re (z) = - \frac{1}{2} \Rightarrow a - b = - 1

.

Counting integer pairs with

a, b \in [- 3, 3]

$a - b = 2$ : $b \in {- 3, - 2, - 1, 0, 1}$ gives $5$ pairs, but $(a, b) = (1, - 1)$ has $a + b = 0$ and is excluded, leaving $4$ pairs.
$a - b = - 1$ : $b \in {- 2, - 1, 0, 1, 2, 3}$ gives $6$ pairs, none with $a + b = 0$ .

Total = 4 + 6 = 10

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Integer answer