If a and b are real numbers between 0 and 1 s | vedclass

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(B) Since the triangle with vertices $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ is equilateral,we have the condition $z_1^2 + z_2^2 + z_3^2 = z_1 z_2

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$a = b = 2 - \sqrt{3}$

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(B) Since the triangle with vertices $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ is equilateral,we have the condition $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.Substituting the values,we get $(a + i)^2 + (1 + bi)^2 + 0 = (a + i)(1 + bi)$.Expanding both sides: $(a^2 - 1 + 2ai) + (1 - b^2 + 2bi) = a + abi + i - b$.Simplifying: $(a^2 - b^2) + 2i(a + b) = (a - b) + i(1 + ab)$.Equating real and imaginary parts:$a^2 - b^2 = a - b$ ... $(i)$$2(a + b) = 1 + ab$ ... $(ii)$From $(i)$,$(a - b)(a + b) = (a - b)$,which implies $(a - b)(a + b - 1) = 0$.So,either $a = b$ or $a + b = 1$.Case $1$: If $a = b$,then from $(ii)$,$2(2a) = 1 + a^2$,so $a^2 - 4a + 1 = 0$.Solving for $a$,$a = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.Since $0 Case $2$: If $a + b = 1$,then $b = 1 - a$. Substituting into $(ii)$,$2(1) = 1 + a(1 - a)$,which gives $a^2 - a + 1 = 0$. This equation has no real roots.Thus,the only solution is $a = b = 2 - \sqrt{3}$.

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $z_1 = a + i$,$z_2 = 1 + bi$,and $z_3 = 0$ form an equilateral triangle,then

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