Let integers a, b in [-3, 3]$a, b \in [-3, 3]$ be such that a + b ≠ 0$a + b \neq 0$. Then the number of all possible ordered pairs (a, b)$(a, b)$, for which | (z - a)/(z + b) | = 1$\left| \frac{z - a}{z + b} \right| = 1$ and | arraycccz + 1 & ω & ω² ω & z + ω² & 1 ω² & 1 & z + ω array | = 1, z in C$\left| \begin{array}{ccc}z + 1 & \omega & \omega^2\\ \omega & z + \omega^2 & 1\\ \omega^2 & 1 & z + \omega \end{array} \right| = 1, z \in \mathbb{C}$, where ω$\omega$ and ω²$\omega^2$ are the roots of x² + x + 1 = 0$x^{2} + x + 1 = 0$, is equal to

Core Logic

This implies (x)/(2) = x x = 0$\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so α = 1$\alpha = 1$.

Step 1: Simplify complex number z

Thus, z = -π ( (1-√(π)i)(√(π)-i)π + 1 + (√(π)-i)(1-√(π)i)1 + π )$z = -\pi \left( \frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i)}{\pi + 1} + \frac{(\sqrt{\pi}-i)(1-\sqrt{\pi}i)}{1 + \pi} \right)$

Step 2: Simplify Bracket

Let's expand the terms directly: z = (π)/(4)(-4) [ √(π) - π i - i - √(π)π + 1 + √(π) - i - π i - √(π)1 + π ]$z = \frac{\pi}{4}(-4) \left[ \frac{\sqrt{\pi} - \pi i - i - \sqrt{\pi}}{\pi + 1} + \frac{\sqrt{\pi} - i - \pi i - \sqrt{\pi}}{1 + \pi} \right]$

Step 3: Find beta

For z = 2π i$z = 2\pi i$: |z| = 2π$|z| = 2\pi$ and (z) = (π)/(2)$\arg(z) = \frac{\pi}{2}$.

Step 4: Distance from Line

Distance of point (α, β) = (1, 4)$(\alpha, \beta) = (1, 4)$ from the line 4x - 3y - 7 = 0$4x - 3y - 7 = 0$:

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines