Question

describe the set of points z in the complex plane that satisfy the given equation; |z-2|=Re(z).

Answer 1

Answer:

Parabola with vertex at point (1,0) that goes to the right (see attached diagram).

Step-by-step explanation:

Let the complex number z be

$z=x+iy,$

then

$z-2=x+iy-2=(x-2)+iy\\ \\Re\ (z-2)=x-2\\ \\Im\ (z-2)=y\\ \\|z-2|=\sqrt{Re^2\ (z-2)+Im^2\ (z-2)}=\sqrt{(x-2)^2+y^2}$

and

$Re\ z=x$

Thus,

$\sqrt{(x-2)^2+y^2}=x$

Square it:

$(x-2)^2+y^2=x^2\\ \\x^2-4x+4+y^2=x^2\\ \\-4x+4+y^2=0\\ \\y^2=4x-4\\ \\y^2=4(x-1)$

This is the equation of parabola with vertex at point (1,0) that goes to the right.