# Thread: complex numbers: |z+1| + |z-1| = 3

1. ## [SOLVED] complex numbers: |z+1| + |z-1| = 3

How can I solve an equation like this:
$\displaystyle |z+1| + |z-1| = 3, z \in \mathbb C$

I've tried expanding z as
$\displaystyle z = x + iy; x,y \in \mathbb R$
and computing the equation
$\displaystyle \sqrt{(x+1)^2 + y^2} + \sqrt{(x-1)^2 + y^2} = 3$
but this way it quickly gets terribly complicated. Plus I have two variables in one expression.

Any other ideas?

2. Originally Posted by lampak
How can I solve an equation like this:
$\displaystyle |z+1| + |z-1| = 3, z \in \mathbb C$

I've tried expanding z as
$\displaystyle z = x + iy; x,y \in \mathbb R$
and computing the equation
$\displaystyle \sqrt{(x+1)^2 + y^2} + \sqrt{(x-1)^2 + y^2} = 3$
but this way it quickly gets terribly complicated. Plus I have two variables in one expression.

Any other ideas?
Do you think there will be a finite number of solutions?

3. Ah, no, there won't. I've forgotten that the task was "mark in the complex plane all numbers satisfying following condition".

So maybe I don't need to calculate anything and just draw? But what? I have a feeling it will be an ellipse but I've never learnt about it.

4. Originally Posted by lampak
Ah, no, there won't. I've forgotten that the task was "mark in the complex plane all numbers satisfying following condition".

So maybe I don't need to calculate anything and just draw? But what? I have a feeling it will be an ellipse but I've never learnt about it.
It is indeed an ellipse.

5. Originally Posted by lampak
How can I solve an equation like this:
$\displaystyle |z+1| + |z-1| = 3, z \in \mathbb C$
This is an ellipse in the complex plane: $\displaystyle \text{Re}^2(z)+2\text{Im}^2(z)=1$.

6. Thanks
But why, actually?

7. Originally Posted by lampak
Thanks
But why, actually?
If you want to avoid doing the algebra, then you can recognise it's an ellipse by noting that:

1. geometrically, |z + 1| is the distance of z from -1.

2. geometrically, |z - 1| is the distance of z from 1.

3. geometrically, |z + 1| + |z - 1| = 3 means (distance of z from -1) + (distance of z from 1) = 3.

4. The above is a well known locus definition of the ellipse (the foci are at z = 1 and z = -1). The cartesian of the ellipse is easily constructed from this: $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $\displaystyle 2a = 3$ and $\displaystyle 1^2 + b^2 = \left(\frac{3}{2}\right)^2$ ....

8. Thanks