If z_1, z_2, z_3 in mathbbC$z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle, whose centroid is z_0$z_0$, then sum_k=1^3 (z_k - z_0)^2$\sum_{k=1}^{3} (z_k - z_0)^2$ is equal to

### Core Logic |1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x$$|1 - i|^x = 2^x \implies (\sqrt{2})^x = 2^x \implies 2^{x/2} = 2^x$$ This implies fracx2 = x implies x = 0$\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so alpha = 1$\alpha = 1$. ### Step 1: Simplify complex number z (1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4$$(1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4$$ Thus, z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right)$z = -\pi \left( \frac{(1-\sqrt{\pi}i)(\sqrt{\pi}-i)}{\pi + 1} + \frac{(\sqrt{\pi}-i)(1-\sqrt{\pi}i)}{1 + \pi} \right)$ ### Step 2: Simplify Bracket Let's expand the terms directly: z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right]$z = \frac{\pi}{4}(-4) \left[ \frac{\sqrt{\pi} - \pi i - i - \sqrt{\pi}}{\pi + 1} + \frac{\sqrt{\pi} - i - \pi i - \sqrt{\pi}}{1 + \pi} \right]$ = -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right]$$= -\pi \left[ \frac{-i(\pi+1)}{\pi+1} + \frac{-i(\pi+1)}{\pi+1} \right]$$ = -pi [ -i - i ] = 2pi i$$= -\pi [ -i - i ] = 2\pi i$$ ### Step 3: Find beta For z = 2pi i$z = 2\pi i$: |z| = 2pi$|z| = 2\pi$ and arg(z) = fracpi2$\arg(z) = \frac{\pi}{2}$. beta = frac|z|arg(z) = frac2pipi/2 = 4$$\beta = \frac{|z|}{\arg(z)} = \frac{2\pi}{\pi/2} = 4$$ ### Step 4: Distance from Line Distance of point (alpha, beta) = (1, 4)$(\alpha, \beta) = (1, 4)$ from the line 4x - 3y - 7 = 0$4x - 3y - 7 = 0$: D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3$$D = \frac{|4(1) - 3(4) - 7|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 - 12 - 7|}{5} = \frac{|-15|}{5} = 3$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines