Question

Let a and b be two two real numbers lying between 0 and 1 such that the points z₁ = a + i, z₂ = 1 + bi and z₃ = 0 form an equilateral triangle, then (a, b) is equal to –

• A. ($\sqrt{3}$/2, $\sqrt{3}$/2)
• B. (2 –$\sqrt{3}$, $\sqrt{3}$/2)
• C. ($\sqrt{3}$/2, 2 –$\sqrt{3}$)
• D. (2 –$\sqrt{3}$, 2 –$\sqrt{3}$)

Answer 1

Answer: (D)

(2 –$\sqrt{3}$, 2 –$\sqrt{3}$)

Answer 2

Sol. Since, z₂, z₃ are the vertices of an equilateral triangle,

|z₁ – z₂| = |z₂ – z₃| = |z₁ – z₃|

Ž |z₁ – z₂|² = |z₂ – z₃|² = |z₁ – z₃|²

Ž (a – 1)² + (1 – b)² = (0 – 1)² + b² = (a – 0)² + (1 – 0)²

Ž (a – 1)² + (1 – b)² = 1 + b² = a² + 1

Now, 1 + b² = a² + 1 Ž b = a[\ a, b > 0]

Thus, (a – 1)² + (1

– a)² = 1 + a²

Ž 2(a² + 1 – 2a) = 1 + a² Ž a² – 4a + 1 = 0

Ž a = $\frac{4 \pm \sqrt{16 - 4}}{2}$ = 2 ±$\sqrt{3}$

As 0 < a < 1, we get

a = b = 2 –$\sqrt{3}$.