Find the set of points in $\mathbb{C}$ s.t. $|z| = Rez + 1$ I was given the above problem as a homework question and I've arrived at a solution but I am not sure if my approach is the best one or even a valid one. Any help would be great!
Consider $z = a + bi : a, b \in \mathbb{R}$ with $a = Rez, b = Imz$. Then we would like $|z| = \sqrt{a^2 + b^2} = a + 1 = Rez + 1.$ To arrive at this, we require $|z| = \sqrt{(a+1)^2} = \sqrt{a^2 + 2a + 1}$, which means $b = \sqrt{2a+1}$ so that $|z| = \sqrt{a^2 + b^2} = \sqrt{a^2 + \sqrt{2a+1}^2} = \sqrt{a^2 + 2a + 1} = \sqrt{(a+1)^2} = a + 1 = Rez + 1$. We also require $2a+1 > 0$ or $a > \frac{-1}{2}.$
This gives us the set $\{a + bi : a > \frac{-1}{2}, b = \sqrt{2a + 1}\}$  

Note that $\sqrt{a^2 + b^2} = \sqrt{a^2 + 2a + 1} \implies \color{red}{b^2 = 2a + 1} \implies b=\color{red}{\pm} \sqrt{2a+1}$.
