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If arg $\left ( z-1 \right )$ = $\large\frac{\pi}{6}$ and arg $\left ( z+1 \right )$ = 2$\large\frac{\pi}{3}$ then prove that $\left | z \right |=1$

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  • If $z=x+iy$ is written in exponential form as $z=r(\cos \theta+i\sin \theta),r=\sqrt{x^2+y^2}$ and the argument $\theta$ is given by the following rule
  • $\theta=\pi-\alpha\Rightarrow \theta=\alpha$
  • $\theta=-\pi+\alpha\Rightarrow \theta=-\alpha$
  • Where $\alpha=\tan^{-1}\mid\large\frac{y}{x}\mid$ and $(x,y)$ lies in one of the four quadrants (or the axes).
Step 1:
Let $z=x+iy\Rightarrow z-1=x+iy-1=(x-1)+iy$
Therefore $\large\frac{y}{x-1}$$=\tan\large\frac{\pi}{6}$
(i.e)$\large\frac{y}{x-1}=\large\frac{1}{\sqrt 3}$
$\Rightarrow x-\sqrt 3y=1$------(1)
(i.e)$\large\frac{y}{x+1}$$=-\sqrt 3$
$\Rightarrow\sqrt 3 x+y=-\sqrt 3$------(2)
Step 2:
Squaring and adding (1) and (2) we get
$(x-\sqrt 3y)^2+(\sqrt 3x+y)^2=1+3$
$x^2+3y^2-2\sqrt 3xy+3x^2+y^2+2\sqrt 3xy=4$
Therefore $x^2+y^2=1=\mid z\mid^2$
$\Rightarrow \mid z\mid=1$
Hence proved.
answered Jun 10, 2013 by sreemathi.v

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