How to prove that $|z+1|^2 + |z-1|^2 = 4$ implies $|z|=1$?

Question

If $z=x-iy$ and $|z+1|^2 + |z-1|^2 = 4$. Then prove that $x^2 + y^2 = 1$.

I have tried $(a+b)^2$ method but I am not able to solve this

Answer 1

With $$z=x-iy$$ we get $$(x+1)^2+y^2+(x-1)^2+y^2=4$$ $$|z+1|^2=|x+1-iy|^2=(x+1)^2+y^2$$

Answer 2

Since $|z+1|^2 + |z-1|^2 = 4$, it follows that $$\begin{align} 4|z|^2=|(z+1)+(z-1)|^2&=|z+1|^2 + |z-1|^2 +2\text{Re}((z+1)(\overline{z-1}))\\&=4+2(|z|^2-1)=2+2|z|^2\end{align}$$ which implies that $x^2+y^2=|z|^2=1$.

Answer 3

If you plug in $z=x-iy$ into the given equation, you get

\begin{align*} 4&=|(x+1)-iy|^2+|(x-1)-iy|^2\\ &=(x+1)^2+y^2+(x-1)^2+y^2\\ &= x^2+2x+1+x^2-2x+1+2y^2\\ &= 2(x^2+y^2)+2, \end{align*} so $2(x^2+y^2)=2$, i.e., $x^2+y^2=1$.

Answer 4

$$4=|z+1|^2 + |z-1|^2 = (z+1)(\bar{z}+1)+(z-1)(\bar{z}-1)=z\bar{z}+z+\bar{z}+1+z\bar{z}-z-\bar{z}+1\\=2|z|^2+2$$

This reduces to $$|z|^2=1$$