Name: Geometric View of Complex Number
Uploaded: 2019-07-19
Duration: 10 min 19 s
Description: Complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis. Watch the video to l...

Question

Using distance formula, prove that the points

1, \frac{- 1 + \sqrt{3} i}{2} and \frac{- 1 - \sqrt{3} i}{2}

form the vertices of an equilateral triangle in complex plane.

Accepted Answer

Let $A (z_{1}) = 1, B (z_{2}) = \frac{- 1 + \sqrt{3} i}{2} and C (z_{3}) = \frac{- 1 - \sqrt{3} i}{2}$

We know that distance between two complex numbers $z_{1}, z_{2} is |z_{2} - z_{1}|$

$∴ A B = |z_{2} - z_{1}| = |\frac{- 1 + \sqrt{3} i}{2} - 1|$

$\Rightarrow A B = |\frac{- 1 + \sqrt{3} i - 2}{2}|$

$\Rightarrow A B = |\frac{- 3 + \sqrt{3} i}{2}|$

$\Rightarrow A B = \sqrt{\frac{9}{4} + \frac{3}{4}}$

$\Rightarrow A B = \sqrt{3} \to (1)$

Similarly, $B C = |\frac{- 1 - \sqrt{3} i}{2} - \frac{- 1 + \sqrt{3} i}{2}|$

$\Rightarrow B C = |\frac{- 1 - \sqrt{3} i}{2} + \frac{1 - \sqrt{3} i}{2}|$

$\Rightarrow B C = |\frac{2 \sqrt{3} i}{2}|$

$\Rightarrow B C = \sqrt{3} \to (2)$ and $C A = \sqrt{3} \to (3)$

From $(1), (2) and (3)$ , we have $A B = B C = C A$

We know that, in an equilateral triangle length of three sides are equal.

Hence,proved.

M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 2: EXERCISE 5.2

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 2: EXERCISE 5.2

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