Points that satisfy the equation $|z + 3| + |z + 1| = 4$

Question

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?

I tried turning it into an ellipse equation $$\frac{\sqrt{(x+3)^2 + y^2}}{2^2}+\frac{\sqrt{(x+1)^2+y^2}}{2^2}=1$$but I don't know what to do from here. — Abdulrahman, Oct 19 '22 at 11:11
"In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant." https://en.wikipedia.org/wiki/Ellipse — bof, Oct 19 '22 at 11:13
In words, your equation says "the distance from $z$ to $-3$ plus the distance from $z$ to $-1$ equals $4$." — bof, Oct 19 '22 at 11:16
Let's pose $\forall z,z' \in \mathbb{C}, d(z,z'):=|z-z'|$ (distance). You look for $\xi:={z\in \mathbb{C}:d(z,-3)+d(z,-1)=4}$, which is the ellipse defined by sum of distances to foci $-3$ and $-1$. — Stéphane Jaouen, Oct 19 '22 at 11:17
@StéphaneJaouen Okay, I understand. Wasn't sure what it was supposed to translate to. — Abdulrahman, Oct 19 '22 at 11:21
I showed here how to convert the square root equation to cartesian equation https://math.stackexchange.com/a/4015746/399263 — zwim, Oct 19 '22 at 11:59
I think the short answer to your question is this. From the "ellipse" hints you know what form the final answer should be in, i.e. the general form of the equation of an ellipse, with various constants to be determined. But now, to determine the numerical values of those various constants, there's just some messy algebra to be done. — Lee Mosher, Oct 19 '22 at 12:04

score 5 · Answer 1 · answered Oct 19 '22 at 12:52

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, $\xi=\{z=x+iy\in\mathbb{C}:\frac{(x+2)^2}{4}+\frac{y^2}{3}=1\}=\{(x,y)\in\mathbb{R}^2:3(x+2)^2+4y^2=12\}.$

score 3 · Answer 2 · answered Oct 19 '22 at 11:15

3

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.

answered Oct 19 '22 at 11:15

cxqntnt

121

Oscar Lanzi · Answer 3 · 2022-10-19T13:41:58.513

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 $(-2,0)$.

Points that satisfy the equation $|z + 3| + |z + 1| = 4$

3 Answers3