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$P$ represents the variable complex number $z$.Find the locus of $P$,if $arg\bigg(\large\frac{z-1}{z+3}\bigg)=\large\frac{\pi}{2}$

This is the fifth part of the multi-part Q8.

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  • If $z=a+ib$ then $\bar{z}=a-ib$.
  • $\mid z\mid=\sqrt{a^2+b^2}$
  • $z^{-1}=\large\frac{a-ib}{a^2+b^2}$
  • $z\bar{z}=a^2+b^2$
  • Also $Re(z)=a,Im(z)=b$
  • Exponential form of a complex number is $z=re^{\large i\theta}=r[\cos\theta+i\sin \theta]$
  • Where $r$ is the modulus and $\theta$ the argument.
  • If $z_1=r_1e^{\large i\theta_1}$ and $z_2=r_2e^{\large i\theta_2}$ are equal,then $r_1=r_2$,$\theta_1=\theta_2$
Step 1:
$P$ is the point representing the variable complex number $z$ such that $arg\big(\large\frac{z-1}{z+3}\big)=\large\frac{\pi}{2}$
Let $z=x+iy\Rightarrow arg\big(\large\frac{z-1}{z+3}\big)=\large\frac{\pi}{2}$
$\quad\quad=\large\frac{(x-1)+iy}{(x+3)+iy}\times \bigg(\large\frac{(x+3)-iy}{(x+3)-iy}\bigg)$
Step 2:
Since $arg\big(\large\frac{z-1}{z+3}\big)=\large\frac{\pi}{2}$
$\Rightarrow \tan^{-1}\large\frac{y(x+3)-y(x-1)}{(x-1)(x+3)+y^2}=\large\frac{\pi}{2}$
$\Rightarrow (x-1)(x+3)+y^2=0$
Step 3:
The locus is a circle with centre $(-1,0)$
$\qquad\;\;=\sqrt 4$
The locus is a circle with centre $(-1,0)$ and radius 2 units.
answered Jun 10, 2013 by sreemathi.v

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