SATHEE: Complex Numbers 4 Question 6

Complex Numbers 4 Question 6

6.

The complex numbers $z_{1}, z_{2}$ and $z_{3}$ satisfying $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1 - i \sqrt{3}}{2}$ are the vertices of a triangle which is

(a) of area zero

(b) right angled isosceles

(d) obtuse angled isosceles

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Answer:

Correct Answer: 6. (c)

Solution:

$\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1 - i \sqrt{3}}{2} = \frac{(1 - i \sqrt{3}) (1 + i \sqrt{3})}{2 (1 + i \sqrt{3})}$

$= \frac{1 - i^{2} 3}{2 (1 + i \sqrt{3})}$

$= \frac{4}{2 (1 + i \sqrt{3})}$

$= \frac{2}{(1 + i \sqrt{3})}$

$\Rightarrow \frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{1 + i \sqrt{3}}{2} = \cos \frac{π}{3} + i \sin \frac{π}{3}$

$\Rightarrow \frac{z_{2} - z_{3}}{z_{1} - z_{3}} = 1$ and $\arg \frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{π}{3}$

Hence, the triangle is an equilateral.

Alternate Solution

$\begin{array}{ll} ∴ \frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{1 - i \sqrt{3}}{2} \\ \Rightarrow \frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{2}{1 - i \sqrt{3}} = \frac{1 + i \sqrt{3}}{2} = \cos \frac{π}{3} + i \sin \frac{π}{3} \\ \Rightarrow \arg \frac{z_{2} - z_{3}}{z_{1} - z_{3}} = \frac{π}{3} and also | \frac{z_{2} - z_{3}}{z_{1} - z_{3}} | = 1 \end{array}$

Therefore, triangle is equilateral.

Complex Numbers 4 Question 6

6.

Answer:

Engineering Preparation

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Complex Numbers 4 Question 6

6.

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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