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Q)

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

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

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A)
Toolbox:
  • 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 $Re(z)=a,Im(z)=b$
  • If $z_1=a+ib,z_2=c+id$
  • $z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$
  • $\mid z_1z_2\mid=\mid z_1\mid\mid z_2\mid$
Step 1:
$P$ represents the variable complex number z.
Let $z=x+iy$
$Re\bigg(\large\frac{z-1}{z+i}\bigg)$$=1$
$Re\bigg(\large\frac{x+iy-1}{x+iy+i}\bigg)$$=1$
$Re\bigg(\large\frac{x-1+iy}{x+i(y+1)}\bigg)$$=1$
Step 2:
$\bigg(\large\frac{x-1+iy}{x+i(y+1)}\bigg)=\bigg(\large\frac{(x-1)+iy}{x+i(y+1)}\bigg)\times\bigg( \large\frac{x-i(y+1)}{x-i(y+1)}\bigg)$
$\qquad\qquad\;\;\;\;\;\;=\bigg(\large\frac{(x-1)x+y(y+1)+i(xy-(x-1)(y+1)}{x^2+(y+1)^2}\bigg)$
Step 3:
The real part of the expression is equal to 1
Therefore $\large\frac{(x-1)x+y(y+1)}{x^2+(y+1)^2}$$=1$
$x^2-x+y^2+y=x^2+y^2+2y+1$
$x+y+1=0$
The locus is a straight line.
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