A Locus in the complex plane

I am having trouble understanding exactly what the following set represents:

The set of all complex numbers $z\neq 5$ such that $\frac{\lvert z -3\rvert }{\lvert z-5\rvert}=\frac{\sqrt2}{2}$

I have a feeling that it represents a circle of some sort, but I do not know how to prove it. I have tried expanding both sides with $z=x+yi$ but did not succeed.

We can rewrite this as

$$|z-3|=\frac{\sqrt 2}2|z-5|$$

which can be interpreted geometrically as the locus of all points that has a distance from the point $3+0i$ that is $\frac{\sqrt 2}2$ times the distance from the point $5+0i$. From geometry we know that is a circle.

If you can solve this geometrically, you can find the nearest point to $3+0i$ and the farthest point from $3+0i$. These are points on the real axis and are easy to find. These points are the diameter of the desired circle, so find the center and radius and you are done.

Your approach of using $z=x+yi$ should also succeed. Use the equation I used, replace the absolute values with square roots of sums of squares, and square both sides of the equation. When you simplify you will get the equation of a circle.

$$|x+yi-3|=\frac{\sqrt 2}2|x+yi-5|$$ $$|(x-3)+yi|=\frac{\sqrt 2}2|(x-5)+yi|$$ $$\sqrt{(x-3)^2+y^2}=\frac{\sqrt 2}2\sqrt{(x-5)^2+y^2}$$ $$(x-3)^2+y^2=\frac 12\left[(x-5)^2+y^2\right]$$ $$2(x-3)^2+2y^2=(x-5)^2+y^2$$ $$2x^2-12x+18+2y^2=x^2-10x+25+y^2$$ $$x^2-2x+y^2=7$$ $$x^2-2x+1+y^2=7+1$$ $$(x-1)^2+y^2=8$$ $$(x-1)^2+y^2=(2\sqrt 2)^2$$

• Thank you very much for your detailed answer! It solves a lot of my confusion – john melon Jan 2 '16 at 21:58