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For any complex number $z$, $arg$ $z$ defines its phase, chosen to be in the interval $0\leq arg z < 360^{∘}$. If $z_{1}, z_{2}$ and $z_{3}$ are three complex numbers with the same modulus but different phases ($arg z_{3} < arg z_{2} < arg z_{1} < 180^{∘}$), then the quantity

$$\frac{arg \left(z_{1}/z_{2}\right)}{arg \left[(z_{1}-z_{3})/(z_{2}-z_{3})\right]}$$

is a constant, and has the value

  1. 2
  2. $\frac{1}{3}$
  3. 1
  4. 3
  5. $\frac{1}{2}$
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$\text{ Suppose, phases of $z_1,z_2$ and $z_3$ are $\theta_1, \theta_2$ and $\theta_3$ respectively}$

$\text{and given: $|z_1| = |z_2|= |z_3| = r$}$

$\text{$\therefore z_1 = r e^{i\theta_1}, z_2 = r e^{i\theta_2}$ and $z_2 = r e^{i\theta_3}$ }$

$\text{$\arg (\frac{z_1}{z_2}) = \arg(\frac{re^{i\theta_1}}{re^{i\theta_2}}) = \arg(e^{i(\theta_1 – \theta_2)})= \theta_1 – \theta_2$ }$

$\text{$\arg(\frac{z_1-z_3}{z_2 – z_3})  = \arg \left(\frac{\frac{z_1}{z_3} – 1}{\frac{z_2}{z_3} – 1}\right) = \arg \left( \frac{e^{i(\theta_1 – \theta_3) } – 1}{e^{i(\theta_2 – \theta_3)} – 1} \right) = \arg \left( \frac{(\cos(\theta_1 – \theta_3)-1) + i\sin(\theta_1 – \theta_3)}{(\cos(\theta_2 – \theta_3) – 1) + i\sin(\theta_2 – \theta_3)} \right) $}$

$\text{ $= \arg \left ( \frac{-2\sin^2(\frac{\theta_1-\theta_3}{2}) + i \sin (\theta_1 – \theta_3)}{-2\sin^2(\frac{\theta_2-\theta_3}{2}) + i \sin (\theta_2 – \theta_3)}\right) $}$

$\text{ $ = \arg (-2\sin^2(\frac{\theta_1-\theta_3}{2}) + i \sin (\theta_1 – \theta_3)) – \arg (-2\sin^2(\frac{\theta_2-\theta_3}{2}) + i \sin (\theta_2 – \theta_3)) $ }$

$\text{$\because -2\sin^2(\frac{\theta_{\{1,2\}}-\theta_3}{2}) < 0$ and $\sin (\theta_{\{1,2\}} – \theta_3) \geq 0$}$

$\text{$\therefore = \pi + \tan^{-1}\left(\frac{\sin (\theta_1 – \theta_3)}{-2\sin^2(\frac{\theta_1-\theta_3}{2})}\right) – \pi  – \tan^{-1}\left(\frac{\sin (\theta_2 – \theta_3)}{-2\sin^2(\frac{\theta_2-\theta_3}{2})}\right) $}$

$\text{$= \tan^{-1} \left(\cot \frac{\theta_3 – \theta_1}{2} \right) – \tan^{-1} \left(\cot \frac{\theta_3 – \theta_2}{2} \right)$}$

$\text{$= \frac{\pi}{2} – \cot^{-1}\left(\cot \frac{\theta_3 – \theta_1}{2}\right) – \frac{\pi}{2} + \cot^{-1}\left(\cot \frac{\theta_3 – \theta_2}{2}\right)$}$

$\text{$= -(\frac{\theta_3 – \theta_1}{2}) + \frac{\theta_3 – \theta_2}{2} = \frac{\theta_1 – \theta_2}{2}$}$

$\text{Therefore, $\arg (\frac{z_1}{z_2}) =  (\theta_1 – \theta_2) $ and $\arg(\frac{z_1-z_3}{z_2 – z_3}) = \frac{(\theta_1 – \theta_2)}{2}$}$

$\textbf{Answer: A }$
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