Geometrical interpretation of a diagram in Complex Analysis I am currently self-studying complex analysis in preparation for my course that will commence next year and I am stuck with this question as shown below:
In the complex plane, draw a picture of $$S=\{z \in \mathbb{C} : \vert z-1 \vert + \vert z+1 \vert = 2\}.$$
I was told that by squaring both sides, the curve is equivalent to $$\vert z^{2} \vert + \vert z^{2} -1 \vert = 1$$ but I cannot work out as such. Can anyone show me how the squaring of both sides resulted into the equation? 
 A: The trick is the identity $|z|^2 = z\bar z$. Squaring both sides of your equation,
$$|z-1|^2 + 2|z-1||z+1| + |z+1|^2 = 4.$$
Now rearrange this and use the identity above to get
$$(z-1)(\overline{z-1}) + (z+1)(\overline{z+1}) + 2|z^2-1| = 4.$$
But $\overline{z-1} = \bar z-1$, so $(z-1)(\overline{z-1}) = (z-1)(\bar z-1) = z\bar z-z-\bar z+1=|z|^2-z-\bar z+1.$ Similarly, $(z+1)(\overline{z+1}) = |z|^2+z+\bar z+1$, so our equation becomes
$$|z|^2-z-\bar z+1 + |z|^2+z+\bar z+1 + 2|z^2-1|=4.$$
Simplifying this gives you the result you're after.
Now, having said that, note that you don't actually need to do any of this. The original equation describes those points whose total distance from $z=1$ and $z=-1$ is equal to $2$. In other words, this is an ellipse with foci at $1$ and $-1$, and the distance is such that the ellipse is degenerate (i.e., it consists of the line segment between the foci).
A: Sum of distances is 2. And inter-focal distance $ 2 a $ also 2. So it is a degenerate ellipse (or parabola to a scale). If the latter is > 2 an ellipse can be formed with foci at $( \pm 1, 0 )$
