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Q.
If $a$ and $b$ are real numbers between 0 and 1 such that the points $z_1=a+i, z_2=1+b i$ and $z_3=0$ form an equilateral triangle, then
Complex Numbers and Quadratic Equations
Solution:
Solution: By the hypothesis $0< a, b< 1$
and $\left|z_1-z_2\right|=\left|z_2-z_3\right|=\left|z_3-z_1\right|$
$\Rightarrow |(a-1)+i(1-b)|=|1+i b|=|a+i| $
$\Rightarrow (a-1)^2+(1-b)^2=1+b^2=a^2+1 $
$\Rightarrow a^2-2 a+1-2 b=0 \text { and } b^2=a^2$
As $0< a, b< 1$ and $a^2=b^2$, we get $a=b$.
$\therefore a^2-2 a+1-2 a=0 \Rightarrow a^2-4 a+1=0 $
$\Rightarrow a=2 \pm \sqrt{3} \text { As } 0< a <1, a=2-\sqrt{3} .$