Let S=zin C:|z-1|=1 and (√(2)-1)(z+ z)-i(z- z)=2√(2)$S=\{z\in C:|z-1|=1 \text{ and } (\sqrt{2}-1)(z+\overline{z})-i(z-\overline{z})=2\sqrt{2}\}$. Let z₁, z₂in S$z_{1}, z_{2}\in S$ be such that |z₁|= zin S|z|$|z_{1}|=\max_{z\in S}|z|$ and |z₂|= zin S|z|$|z_{2}|=\min_{z\in S}|z|$. Then |√(2)z₁-z₂|²$|\sqrt{2}z_{1}-z_{2}|^{2}$ equals:

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