My approach was to set $z=x+iy$ and the solve. I got down to
$$ |-x| + |-i(y-1)| = 0, \qquad x = i(y-1) $$
Is this the right approach?? It just doesn't seem right and I'm not really sure how to interpret my answer. Please tell me the correct way to do this.
Hello, thank you for your responses. Here are my steps now: $$ |z-i| = 2|z+i| $$ $$ |x + iy -i| = 2|x + iy +i| $$ $$ |x + i(y -1)| = 2|x + i(y+1)| $$ $$ x^2 + (y-1)^2 = 2(x^2 + (y+1)^2) $$ $$ x^2 + (y^2 -2y +1) = 2(x^2 + (y^2 + 2y + 1)) $$ $$ x^2 + y^2 -2y +1 = 2x^2 + 2y^2 + 4y + 2 $$ $$ x^2 + y^2 + 6y + 1 = 0 $$ Equation of a circle is: $$ (x-h)^2 + (y-k)^2 = r^2 $$ Complete the square to get: $$ x^2 + (y+3)^2 = 8 $$ So it is a circle centered at (0,-3) with radius 2sqrt(2).
Thank you for your help!!
Just a little correction:
$$|x+iy-i|=2|x+iy+i|$$ $$or, x^2+(y-1)^2=4\{x^2+(y+1)^2\}$$ $$or, x^2+y^2-2y+1=4x^2+4y^2+8y+4$$ $$or, 3x^2+3y^2+10y+3=0$$ Which is the equation of a circle.