Plotting the set $\left\{ z\in \mathbb{C} : \left| z+i \right| =2\left| z \right| \right\} $ I have a question which can be simple, though I couldn't catch the idea.
Consider the Set given below:
$$E=\left\{ z\in \mathbb{C} : \left| z+i \right| =2\left| z \right|  \right\} $$
I have to find its graph to plot.
The Answer is the following: the Set $E$ is a circle, centred at $i/3$ and of radius $2/3$
I was not able of visualizing this set, I absurdly thought about a spiral. Any calculations neither didn't lead me to a solution. And especially, my conclusion was that this graph couldn't be a circle at all.
Thanks for helping me in advance.
 A: Hint 
Put $z=x+yi$ in given equation to get $$x^2+(y+1)^2=4(x^2+y^2)$$
$$\Rightarrow 3(x^2+y^2)-2y-1=0$$
$$\Rightarrow x^2+y^2-\frac 23 y-\frac 13=0$$
Which is the equation of required circle with centre $(0,\frac 13)$ and radius $\frac 23$ but since we are dealing in complex plane the centre is $\frac i3$
A: A good way to read an equation like this is geometrically. We want all points such that their distance from $-i$ is two times their distance from $0$. It should be clear that both $i$ and $-\frac13i$ both satisfy this condition. What ever the set looks like, it must be symmetric across the line through $-i$ and $0$, which is to say, the imaginary axis.
To see that the desired set is a circle is a little trickier. The general such problem is the Circles of Appollonius problem, which is linked in a comment under your question. A more modern approach is to follow Descartes, and write down an equation in $x$ and $y$ satisfied by points in the desired set. This equation is also given in a comment under your question.
A: Here is a "non-Cartesian" solution with a complex square completion.
The equation of a circle in the complex plane around point $a$ is:


*

*$|z-a|^2 = (z-a)(\bar z - \bar a)= |z|^2 - 2 Re(\bar a z) + |a|^2 = r^2$


Squaring the given equation:


*

*$|z|^2 - 2 Re(i z) + 1 = 4 |z|^2$


Rearranging and square completing:


*

*$|z|^2 - 2 Re(-\frac{i}{3} z) - \frac{1}{3} = |z - \frac{i}{3} |^2 - \frac{1}{9} - \frac{1}{3}= 0$
$$ \rightarrow |z -\frac{i}{3}| = \frac{2}{3}$$

