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?

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 $(-2,0)$.
