SATHEE: Complex Numbers 4 Question 11

Complex Numbers 4 Question 11

11.

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 $a = \dots$ and $b =$ …

$(1990, 2 M)$

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Answer:

Correct Answer: 11. $a = b = 2 \pm \sqrt{3}$

Solution:

Since, $z_{1}, z_{2}$ and $z_{3}$ form an equilateral triangle.

$\begin{aligned} \Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} \\ \Rightarrow (a + i)^{2} + (1 + i b)^{2} + (0)^{2} = (a + i) (1 + i b) + 0 + 0 \\ \Rightarrow a^{2} - 1 + 2 i a + 1 - b^{2} + 2 i b = a + i (a b + 1) - b \\ \Rightarrow (a^{2} - b^{2}) + 2 i (a + b) = (a - b) + i (a b + 1) \\ \Rightarrow a^{2} - b^{2} = a - b \\ and 2 (a + b) = a b + 1 \\ \Rightarrow (a = b or a + b = 1) \\ and 2 (a + b) = a b + 1 \\ If a = b, 2 (2 a) = a^{2} + 1 \\ \Rightarrow a^{2} - 4 a + 1 = 0 \\ \Rightarrow a = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} \\ If a + b = 1, 2 = a (1 - a) + 1 \Rightarrow a^{2} - a + 1 = 0 \\ \Rightarrow a = \frac{1 \pm \sqrt{1 - 4}}{2}, but a and b \in R \\ ∴ Only solution when a = b \\ a = b = 2 \pm \sqrt{3} \\ a = b = 2 - \sqrt{3} [∵ a, b \in (0, 1)] \end{aligned}$

Complex Numbers 4 Question 11

11.

Answer:

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Complex Numbers 4 Question 11

11.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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