# Thread: Set of complex numbers that satisfy a given equation

1. ## Set of complex numbers that satisfy a given equation

The equation is $|z-a| - |z+a| = 2c$

i) Let $a \in \mathbb{R}$ and $c > 0$ be fixed. Describe the set of points $z$ satisfying the above equation for every possible choice of $a$ and $c$.

ii) Now let $a \in \mathbb{C}$ and, using a rotation of the plane, describe the locus of points satisfying the above equation.

I tried algebra for both i) and ii) and just got a big mess, no matter what form I converted the complex number to, or even just matching up real and imaginary parts.
I am not sure how else to find the set of $z$ that would satisfy the above equation.

I am also lost by what they mean by locus in ii), I am assuming it must be some sort of conic section, either a hyperbola or an ellipse. But I dont know how to show this either.

2. Originally Posted by Sheld
The equation is $|z-a| - |z+a| = 2c$
Think of it this way, $|z-a|$ is the distance from $z$ to $a$.
What is the locus of points whose distances from two points have a constant difference?
The difference in the distance of z to a and the distance from z to -a is equal 2c.

3. Originally Posted by Plato
Think of it this way, $|z-a|$ is the distance from $z$ to $a$.
What is the locus of points whose distances from two points have a constant difference?
The difference in the distance of z to a and the distance from z to -a is equal 2c.
@OP: The above locus is valid for the case a < c. The case a = c will be the degenerate case of the above locus. And a > c needs to be considered in light of the above.

4. For "locus", you can substitute "set of all".

5. So it's a hyperbola when $a < c$?
I guess it would be a parabola when $a = c$? And if $a > c$ it's a ellipse?

And I must find the equation of this hyperbola? $z=$ something to do with $a$ and $c$?

$a$ is the length from the foci to the center and $c$ is the length from the vertex to the center. What would this be centered at though? The origin?

Darn, I wish I remembered the formula for a hyperbola. I am sure $a$ and $c$ are used somehow.

6. The standard rectangular form for an equation of a hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

x and y may be interchanged depending on which way the hyperbola is facing.

7. Wow, I think I may have it.

If $c > a$
Its the set of $z$ s.t. $\frac{x^2}{c^2}-\frac{y^2}{c^2-a^2} = 1$

If $c < a$
Its the set of $z$ s.t. $\frac{x^2}{c^2}+\frac{y^2}{c^2-a^2} = 1$

I am not sure how to write the equation for when $c=a$. It seems like it's two lines intersecting at some point. Not sure what the point of intersection is either.

8. Originally Posted by Sheld
So it's a hyperbola when $a < c$? Mr F says: No. There's no locus because there's no value of z that satisfies the condition. If you take an algebraic approach by substituting z = x + iy into |z - 1| - |z + 1| = 4, say, you might be fooled into thinking that yuo get an ellipse. However, there is an important restriction that arises during the calculation. When you impose this restriction, you end up with the null set.

I guess it would be a parabola when $a = c$? Mr F says: No. It's a ray, the degenerate case of the above ....

And if $a > c$ it's a ellipse? Mr F says: No. It's a single branch of a hyperbola. Which branch will depend on whether c is positive or negative.

And I must find the equation of this hyperbola? $z=$ something to do with $a$ and $c$?

$a$ is the length from the foci to the center and $c$ is the length from the vertex to the center. What would this be centered at though? The origin?

Darn, I wish I remembered the formula for a hyperbola. I am sure $a$ and $c$ are used somehow.
Originally Posted by Sheld
Wow, I think I may have it.

If $c > a$
Its the set of $z$ s.t. $\frac{x^2}{c^2}-\frac{y^2}{c^2-a^2} = 1$

If $c < a$
Its the set of $z$ s.t. $\frac{x^2}{c^2}+\frac{y^2}{c^2-a^2} = 1$

I am not sure how to write the equation for when $c=a$. It seems like it's two lines intersecting at some point. Not sure what the point of intersection is either.
No. See above.

Edit: Fixed a typo (had the first and last cases the wrong way around) which, I don't think, affected the subsequent posts by the OP.

9. But, c is greater than zero. I assuming it'd be the right branch. I am not sure how my equations are wrong. Is there another approach that is not algebraic? Could you explain it please? I've tried to draw pictures of these things, but nothing makes sense.

10. Originally Posted by Sheld
But, c is greater than zero. I assuming it'd be the right branch. I am not sure how my equations are wrong. Is there another approach that is not algebraic? Could you explain it please? I've tried to draw pictures of these things, but nothing makes sense.
OK. Left hand branch of a hyperbola if 0 < c < a.

Ray from z = -a in direction of negative real axis if c = a.

No solution if c > a.

This all follows from the geometric approach suggested in post #2.

I suggest you consider some concrete examples:

a > c: |z - 2| - |z + 2| = 1

a = c: |z - 2| - |z + 2| = 4

a < c: |z - 2| - |z + 2| = 6

If you're uncomfortable with the geometric approach, substitute z = x + iy and take an algebraic approach.

Since you have not shown any working as to how you got your answers, it's impossible to know the mistakes you have made in getting the wrong equations. You will need to show detaield working if your work is to be reviewed properly.

Note: I corrected a typo in my first post.

11. Here is my work for c<a (algebraic), using your example $|z-2| - |z+2| = 1$

$\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2} = 1$

$\sqrt{(x-2)^2+y^2} = \sqrt{(x+2)^2+y^2} =1
$

$x^2-4x+4+y^2 = 2\sqrt{x^2+4x+4+y^2}+x^2+4x+y^2+5$
$-8x-1 = 2\sqrt{x^2+4x+4+y^2}$
$-4x-\frac{1}{2} = \sqrt{x^2+4x+4+y^2}$
$16x^2+4x+\frac{1}{4} = x^2-4x+4+y^2$
$15x^2+8x-\frac{15}{4} = y^2$
$\pm\sqrt{15x^2+8x-\frac{15}{4}} = y$

I am not sure what this tells me about $z = x+iy$ since a can't solve explicitly for $x$

Thanks again for all your help.

12. Originally Posted by Sheld
Here is my work for c<a (algebraic), using your example $|z-2| - |z+2| = 1$

$\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2} = 1$

$\sqrt{(x-2)^2+y^2} = \sqrt{(x+2)^2+y^2} =1
$

$x^2-4x+4+y^2 = 2\sqrt{x^2+4x+4+y^2}+x^2+4x+y^2+5$
$-8x-1 = 2\sqrt{x^2+4x+4+y^2}$
$-4x-\frac{1}{2} = \sqrt{x^2+4x+4+y^2}$ **
$16x^2+4x+\frac{1}{4} = x^2-4x+4+y^2$
$15x^2+8x-\frac{15}{4} = y^2$
$\pm\sqrt{15x^2+8x-\frac{15}{4}} = y$ Mr F says: This step is not required. From the previous step, you now have to re-arrange into the stadard form of a hyperbola.

I am not sure what this tells me about $z = x+iy$ since a can't solve explicitly for $x$

Thanks again for all your help.
I will assume there is no careless mistake in your algebra since you correctly get a hyperbola at the end.

There is an implicit restriction in the line I labelled **, namely $-4x - \frac{1}{2} \geq 0$. This restriction needs to be imposed once you have cartesian equation for the hyperbola. That's why only one branch is defined.

13. Oh so that is where the restriction comes from. I can't believe I didn't see that. Thanks again for your help, I finally understand the problem now!

-Best regards