Proving that $ \mid z+1\mid^{2}=2\mid z\mid^{2} \Leftrightarrow \mid z-1\mid^{2}=2 $ Let z be a complex number.
How does one prove, using the properties of the conjugate and the modulus, that:
$$ \mid z+1\mid^{2}=2\mid z\mid^{2}
\Leftrightarrow
\mid z-1\mid^{2}=2 $$

An elementary solution:
First, note that $|x+iy|^2=x^2+y^2,|x+1+iy|^2=x^2+2x+1+y^2$, and $|z-1  |^2=(x-1)^2+y^2.$  
We have then
$$ x^2+2x+1+y^2=2x^2+2y^2$$
or $$x^2-2x+1-2+y^2=0$$ 
which is equivalent with the following
$$(x-1)^2+y^2=2$$   
and again, this is what we needed
$$|z-1  |^2=2.$$
