Proving that $ \mid z+1\mid^{2}=2\mid z\mid^{2} \Leftrightarrow \mid z-1\mid^{2}=2 $

Question

Let z be a complex number. How does one prove, using the properties of the conjugate and the modulus, that:

$$ \mid z+1\mid^{2}=2\mid z\mid^{2} \Leftrightarrow \mid z-1\mid^{2}=2 $$

Answer 1

The two esential properties that you have to use are $\overline{w_1 + w_2} = \overline{w_1} + \overline{w_2}$ and ${|w|}^2 = w \overline{w}$ for each $w , w_1 , w_2 \in \mathbb{C}$.

Fix $z \in \mathbb{C}$. With these two properties, you should see that (as $z + \overline{z} = 2 Re z$) $$ {|z + 1|}^2 = {|z|}^2 + 2 (Re z) + 1 \qquad \mbox{ and } \qquad {|z - 1|}^2 = {|z|}^2 - 2 (Re z) + 1\mbox{.} $$ Thus, $$ {|z + 1|}^2 = 2 {|z|}^2 \quad \Longleftrightarrow \quad {|z|}^2 = 2 (Re z) + 1 \quad \Longleftrightarrow \quad {|z - 1|}^2 = 2\mbox{.} $$

Answer 2

Hint: Use (three times) the fact that $|z|^2=z.\overline z$.

Answer 3

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

$$|z-1 |^2=2.$$