Let α = -1 + i√(3)2$\alpha = \frac{-1 + \mathrm{i}\sqrt{3}}{2}$ and β = -1 - i√(3)2,i = √(-1)$\beta = \frac{-1 - \mathrm{i}\sqrt{3}}{2},\mathrm{i} = \sqrt{-1}$. If (7 - 7α +9β)²⁰ + (9 + 7α -7β)²⁰ + (-7 + 9α +7β)²⁰ + (14 + 7α +7β)²⁰ = m¹⁰$(7 - 7\alpha +9\beta)^{20} + (9 + 7\alpha -7\beta)^{20} + (-7 + 9\alpha +7\beta)^{20} + (14 + 7\alpha +7\beta)^{20} = \mathrm{m}^{10}$, then m is ____.

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