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Problem : The complex numbers $z_1, z_2,z_3$ are satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ $z_1,z_2,z_3$ are vertices of a triangle , which type of triangle is it :

My approach :

$|\frac{z_1-z_3}{z_2-z_3}|=|\frac{1-i\sqrt{3}}{2}| = 1 $

argument $\frac{z_1-z_3}{z_2-z_3} = \cos\frac{\pi}{3}-\sin\frac{\pi}{3}$ =$\frac{\pi}{3}$

So, I am unable to work out, whether it a isosceles or equilateral triangle. Please suggest will be of great help, thanks.

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  • $\begingroup$ Pick up arbitrary $z_2, z_3$ and solve for $z_1$ $\endgroup$ – uranix Dec 18 '16 at 11:45
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Given $$\frac{z_1-z_3}{z_2-z_3} = \frac{1-i\sqrt{3}}{2} =\frac{1}{2}-i\frac{\sqrt{3}}{2} = \cos(-\frac{\pi}{3}) +i\sin (-\frac{\pi}{3})$$ $$\Rightarrow \frac{z_2-z_3}{z_1-z_3} =\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$$ $$\Rightarrow |\frac{z_2-z_3}{z_1-z_3}| = 1 \text{and arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$$ If $z_1, z_2, z_3$ represent the vertices $A, B, C$ of a triangle ABC, then, $$|\frac{z_2-z_3}{z_1-z_3}| =1 \Rightarrow |z_2-z_3| = |z_1-z_3|$$ This gives us $BC=AC$. Also, $\text{arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$ gives $\angle BCA =\frac{\pi}{3}$. So, ABC is equilateral.

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$$\left|\frac{z_1-z_3}{z_2-z_3}\right|= 1 \Rightarrow |z_1-z_3|=|z_2-z_3|$$

using proportion rule:

$$\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2} \Rightarrow \frac{z_1-z_3-(z_2-z_3)}{z_2-z_3}=\frac{1-i\sqrt{3}-2}{2} \Rightarrow \left|\frac{z_1-z_2}{z_2-z_3}\right|=1 \Rightarrow |z_1-z_2|=|z_2-z_3|$$

So,

$$|z_1-z_3|=|z_2-z_3|=|z_1-z_2|$$

So it is equilateral.

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Hint:

$\implies z_1=(\cdots)z_2-(\cdots)z_3$

$$\dfrac{z_1-z_2}{z_2-z_3}=?$$

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  • $\begingroup$ This answer needs deciphering $\endgroup$ – polfosol Dec 18 '16 at 12:36
  • $\begingroup$ @polfosol, Have you calculated $$\dfrac{z_1-z_2}{z_2-z_3}=?$$ We already have $$|z_1-z_3|=|z_2-z_3|$$ $\endgroup$ – lab bhattacharjee Dec 18 '16 at 12:42

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