Points that satisfy the equation $|z + 3| + |z + 1| = 4$ I am trying to find the points that satisfy the equation $$|z+3|+|z+1|=4$$
Substituting the value of $z$ and evaluating its modulus gives me $$\sqrt{(x+3)^2+y^2}+\sqrt{(x+1)^2+y^2}=4$$
What I tried to do is to square both sides giving me $$a+b+2\sqrt{ab}=16$$ $$a=(x+3)^2+y^2,\ b=(x+1)^2+y^2$$
and I know what follows is going to be a lengthy and time-consuming process. Is there any faster or easier method to solve this problem?
 A: To make it easy, you can define: t = x + 2, then
$$\sqrt{(t+1)^2+y^2}=4 - \sqrt{(t-1)^2+y^2}$$ 
Squared two side two times (always left a square root on the left hand side),
Finally you can get an ellipse equation.
A: Let's pose $\forall z,z' \in \mathbb{C}, d(z,z'):=|z-z'|$(distance). You look for $\xi:=\{z\in \mathbb{C}:d(z,-3)+d(z,-1)=4\}$$\color{red}{(*)}$. You know that $\xi$  is the ellipse defined by sum of distances to foci $F_1=−3$ and $F_2=−1$.
By translation (i.e. by the change of variable $x=-2+t$ consisting in choosing as the origin of the frame the middle $-2$ of $[F_1F_2]$), we are reduced to the search for $a>0$ and $b>0$ such that $\frac{t^2}{a^2}+\frac{y^2}{b^2}=1$.
Coming back to the definition $\color{red}{(*)}$ of $\xi$, and considering two well-chosen points $A$ with abscissa $0$ and B with ordinate $0$, you just have to solve $a+1+a-1=4$ and $2\sqrt{b^2+1}=4$.
Finally, $\xi=\{z=x+iy\in\mathbb{C}:\frac{(x+2)^2}{4}+\frac{y^2}{3}=1\}=\{(x,y)\in\mathbb{R}^2:3(x+2)^2+4y^2=12\}.$

A: How to obtain the equation without getting bogged down in square root radicals:
Start with
$|z+3|+|z+1|=4$
and note that
$|z+3|^2-|z+1|^2=(x+3)^2+y^2-(x+1)^2-y^2=4x+8.$
Thus from the difference of squares factorization we must accept
$|z+3|-|z+1|=(4x+8)/4=x+2.$
So
$|z+3|=(1/2)[(|z+3|+|z+1|)+(|z+3|-|z+1|)]=(4+x+2)/2=(x+6)/2.$
Squaring then gives
$|z+3|^2=(x+3)^2+y^2=(x+6)^2/4$
$4x^2+24x+36+4y^2=x^2+12x+36$
$3x^2+12x+4y^2=0.$
Completing the square in the $x$ variable gives
$3(x+2)^2+4y^2=12,$
which we recognize as a standard form for an ellipse centered at $(-2,0)$.
