Question

brain teaser:
find the area of the region in which the complex number \[z(=x+iy) \] lies, under the following conditions
\[z_1=\frac{ 2z-i }{ iz+2 }~~,|x_1|<1,y_1>0,|z_1|>1 \\\\\\where--\\z_1=x_1+iy_1\]

Answer 1

So \(-1<x_1<1\) and \(y_1>0\)? No absolute value on the \(y\)?

Answer 2

yeah that's all given no more :(

Answer 3

wanna work on this in morning :'(

Answer 4

i was doing it like this--
\[z_1=\frac{ 2z-i }{ iz+2 } \\x_1+iy_1=\frac{ 2(x+iy)-i }{ i(x+y)+2 } \\=\frac{ 2x+i(2y-1) }{ (2-y)+ix }*\frac{ (2-y)-ix }{ (2-y)-ix } \\=\frac{ 2x(2-y)+x(2y-1) }{ (2-y)^2+x^2 }+\frac{ i[(2y-1)(2-y)-2x^2] }{ (2-y)^2 +x^2}\]

Answer 5

@ikram002p :)

Answer 6

Looks like \(z_1\) can be any number in this region.