What is the equation of the circle for |(z+1)/(z-1)|=3?

[trenekas]

Hello! Few weeks ago we started learning complex number. And i have some questions about that because not all i understand and also don't know if my solution is good :)
Also don't know if my name of topic is good :) if something goes wrong just say. I'd be grateful

My task: |(z+1)/(z-1)|=3 need to calculate and graph that.
I know that modulus of complex number z=x+yi is equal to |z|=√x^2+y^2

The same i tried to do in my exercise.
|(z+1)/(z-1)|=3
$\sqrt{(x+1)^2+y^2/(x-1)^2+y^2}=3$ then i sqaured both sides
$(x+1)^2+y^2/(x-1)^2+y^2=9$
$(x+1)^2+y^2=9((x-1)^2+y^2))$
$x^2+2x+1+y^2=9x^2-18x+9+9y^2$
$8x^2-20x+8+8y^2=0$
$x^2+1+y^2=(20x/8)$
$x^2+y^2=(20x/8)-1$
And that's it what i done. For real i have no idea how graph that and not sure if this solution is good. Thanks for advise what i need to do next or correct my mistakes :)

P.S. I think this is a circle, but don't know where is center and what is radius.

 

[Curious3141]

Hello! Few weeks ago we started learning complex number. And i have some questions about that because not all i understand and also don't know if my solution is good :)
Also don't know if my name of topic is good :) if something goes wrong just say. I'd be grateful

My task: |(z+1)/(z-1)|=3 need to calculate and graph that.
I know that modulus of complex number z=x+yi is equal to |z|=√x^2+y^2

The same i tried to do in my exercise.
|(z+1)/(z-1)|=3
$\sqrt{(x+1)^2+y^2/(x-1)^2+y^2}=3$ then i sqaured both sides
$(x+1)^2+y^2/(x-1)^2+y^2=9$
$(x+1)^2+y^2=9((x-1)^2+y^2))$
$x^2+2x+1+y^2=9x^2-18x+9+9y^2$
$8x^2-20x+8+8y^2=0$
$x^2+1+y^2=(20x/8)$
$x^2+y^2=(20x/8)-1$
And that's it what i done. For real i have no idea how graph that and not sure if this solution is good. Thanks for advise what i need to do next or correct my mistakes :)

P.S. I think this is a circle, but don't know where is center and what is radius.

You're on the right track. It is a circle. Keep all the terms on the left hand side (LHS), complete the square for x. Finally, move the constant over to the RHS, then compare with the standard equation for the circle.