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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!

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!

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 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.

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