Algebra> complex no problems!...

1) im[z+1/z-1] = 0

find the locus of z!

2) arg[z-2/z-3]= pie/4

find the locus of z!

Abhijat Pandey , 13 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below
1)
$z = x + iy$
$\frac{z+1}{z-1} = \frac{x+iy+1}{x+iy-1}$
$\frac{z+1}{z-1} = \frac{(x+1)+iy}{(x-1)+iy}$
$\frac{z+1}{z-1} = \frac{(x+1)+iy}{(x-1)+iy}.\frac{(x-1)-iy}{(x-1)-iy}$
$\frac{z+1}{z-1} = \frac{(x+1)+iy}{(x-1)^{2}+y^{2}}.((x-1)-iy)$
$\frac{z+1}{z-1} = \frac{(x^{2}-1-iy(x+1)+iy(x-1))+y^{2})}{(x-1)^{2}+y^{2}}$
$\frac{z+1}{z-1} = \frac{(x^{2}-1+y^{2}+i(-xy-y+xy-y)}{(x-1)^{2}+y^{2}}$
$\frac{z+1}{z-1} = \frac{(x^{2}-1+y^{2}+i(-2y)}{(x-1)^{2}+y^{2}}$
$Im[\frac{z+1}{z-1}] = 0$
$\Rightarrow y = 0$
is the locus.

2)
$\frac{z-2}{z-3} = \frac{(x-2)+iy}{(x-3)+iy}$
$\frac{z-2}{z-3} = \frac{(x-2)+iy}{(x-3)+iy}.\frac{(x-3)-iy}{(x-3)-iy}$
$\frac{z-2}{z-3} = \frac{(x-2)+iy}{(x-3)^{2}+y^{2}}.((x-3)-iy)$
$\frac{z-2}{z-3} = \frac{(x-2)(x-3)-iy(x-2)+iy(x-3)+y^{2}}{(x-3)^{2}+y^{2}}$
$\frac{z-2}{z-3} = \frac{(x-2)(x-3)+y^{2}-iy}{(x-3)^{2}+y^{2}}$
$arg[\frac{z-2}{z-3}] = \frac{\pi }{4}$
$\Rightarrow Re[\frac{z-2}{z-3}] = Im[\frac{z-2}{z-3}]$
$x^{2}-5x+6+y^{2} = -y$
$x^{2}-5x+y+6+y^{2} = 0$
So locus ia s circle.