NVAMediumJEE Main 2025 · 29 January, Shift 2Modulus & Argument

Mathematics Question from JEE Main 2025 · 29 January, Shift 2

Let integers a,b[3,3]a, b \in [-3,3] be such that a+b0a + b \neq 0.

Then the number of all possible ordered pairs (a,b)(a, b), for which

zaz+b=1\left| \frac{z - a}{z + b} \right| = 1

and

z+1ωω2ωz+ω21ω21z+ω=1,  zC,\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 1, \; z \in \mathbb{C},

where ω\omega and ω2\omega^2 are the roots of x2+x+1=0x^2 + x + 1 = 0, is equal to:

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