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

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Yes your way is fine, indeed by $z=x+iy$ we have that

$$3(x-1)^2+3y^2= (x+1)^2+y^2\iff 2x^2-8x+2+2y^2=0$$

$$\iff 2x^2-8x+2+6+2y^2=6\iff 2((x-2)^2+y^2)=6$$

$$\iff 2|z-2|^2=6\iff |z-2|^2=3$$

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Use the property $|z|^2 = z \bar{z}$. $$ 3(z-1)(\bar{z}-1) = (z+1)(\bar{z}+1) \Leftrightarrow 2 (|z|^2 - 2(z+\bar{z}) + 1) = 0 \\\Leftrightarrow |z|^2 - 2(z+\bar{z}) + 1 = 0 \Leftrightarrow |z|^2 - 2(z+\bar{z}) + 4 = 3 \\ \Leftrightarrow |z-2|^2 = 3. $$

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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 $a=2, b=\sqrt3, c=1$

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If $x=\Re z$ then $3|z-1|^{2}=|z+1|^{2}$ iff $3|z|^{2} -6x+3=|z|^{2}+2x+1$ iff $2|z|^{2}-8x+2=0$ iff $|z|^{2}-4x+1=0$ iff $|z|^{2}-4x+4=3$ iff $|z-2|^{2}=3$.

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