Why is this set a circle: $E = \{z \in \mathbb{C} : |z + i| = 2|z|\}$.

Question

I am told

the set $E = \{z \in \mathbb{C} : |z + i| = 2|z|\}$ is a circle centered at $\dfrac{i}{3}$ of radius $\dfrac{2}{3}$.

We can write this equation like:

$$\sqrt{x^2 + (y + i)^2} = 2\sqrt{x^2 + y^2} \implies\\ x^2 + (y+i)^2 = 4x^2 + 4y^2 \implies \\ x^2 + y^2 + 2iy - 1 = 4x^2 + 4y^2 \implies\\ 3x^2 + 3y^2 - 2iy = -1$$

Already, with that $y$ term, this can't be a circle, right?

Answer 1

Here a purely complex approach:

A circle in $\mathbb{C}$ with radius $r$ around $a$ is given by $\boxed{|z-a| = r}$. Squaring gives an equivalent equation:

$$ |z-a|^2 = (z-a)(\bar z - \bar a) = |z|^2 + |a|^2 - (a\bar z + \bar a z) = \boxed{|z|^2 +\color{blue}{|a|^2} - 2\Re{(\color{blue}{\bar a} z)} = r^2}$$

In your case you only need to check whether you can rearrange your given equation into the boxed form. From there you can read off immediately the radius and the center: $$|z + i|^2 = |z - (-i)|^2 = 4|z|^2 \Leftrightarrow |z|^2 + 1 - 2\Re{(iz)}= 4 |z|^2 \Leftrightarrow |z|^2 + \frac{2}{3}\Re{(iz)} = \frac{1}{3}$$ Now, a quick complex square completion leads to the desired result: $$\frac{1}{3} = |z|^2 + \frac{2}{3}\Re{(iz)} = |z|^2 - 2\Re{(\color{blue}{-\frac{1}{3}i}z)} +\color{blue}{\frac{1}{9}} - \frac{1}{9} \Leftrightarrow \boxed{|z|^2 + \color{blue}{\frac{1}{9}} -2\Re{(\color{blue}{-\frac{1}{3}i}z)} = \frac{4}{9} }$$ This describes a circle around $\color{blue}{a = \frac{1}{3}i}$ with radius $r= \frac{2}{3}$.