Prove $3|z-1|^2 = |z+1|^2 \iff |z-2|^2 = 3$ I'm struggling with this prove question. I tried starting with let $z = x + iy$, and substituting the three in the first eqn with $|z-2|^2$, with the intention of eventually equating the LHS with the RHS, and I ended up with something most likely incorrect. 
Any help would be greatly appreciated, the question is as follows:

Prove $$ 3|z-1|^2 = |z+1|^2 \quad \iff \quad|z-2|^2 = 3 $$


Generalization:
Let $$|z-a|=b,z=a+b\cos t+ib\sin t$$  where $a,b,t$ are real
$$\left|\dfrac{z-c}{z+c}\right|=\sqrt{\dfrac{(a-c+b\cos t)^2+b^2\sin^2t}{(a+c+b\cos t)^2+b^2\sin^2t}}=\sqrt{\dfrac{(a-c)^2+b^2+2b(a-c)\cos t}{(a+c)^2+b^2+2b(a+c)\cos t}}$$
which will be $$\left|\dfrac{a-c}{a+c}\right|$$ if
$$\dfrac{2b(a-c)}{2b(a+c)}=\dfrac{(a-c)^2+b^2}{(a+c)^2+b^2}$$
$$\iff\dfrac{a-c}{a+c}=\dfrac{a^2+b^2+c^2-2ca}{a^2+b^2+c^2+2ca}$$
$$\iff\dfrac ac=\dfrac{a^2+b^2+c^2}{2ca}\iff a^2=b^2+c^2$$
Here $a=2, b=\sqrt3, c=1$
