Showing that the solutions form an ellipse

[PsychonautQQ]

Homework Statement

Let |z+1| + |z-1| = 7 where z are complex numbers. Show that the solutions to this equation form an ellipse with foci at (+/-)1

Homework Equations

(x^2 / a) + (y^2 / b) = 1 equation for an ellipse

The Attempt at a Solution

I set z = a + bi and so |z-1| = ((a-1)^2 + b^2)^1/2 and analogously for |z+1|. I'm a bit confused now, if I square both sides I wind up with a big mess, how to I beat this into a form that's recognizable as an ellipse?

 

[Ray Vickson]

Homework Statement

Let |z+1| + |z-1| = 7 where z are complex numbers. Show that the solutions to this equation form an ellipse with foci at (+/-)1

Homework Equations

(x^2 / a) + (y^2 / b) = 1 equation for an ellipse

The Attempt at a Solution

I set z = a + bi and so |z-1| = ((a-1)^2 + b^2)^1/2 and analogously for |z+1|. I'm a bit confused now, if I square both sides I wind up with a big mess, how to I beat this into a form that's recognizable as an ellipse?

Keep expanding: if $d_1=|z-i|$ and $d_2 = |z+i|$, you have $d_1+d_2 = 7$, so $2 d_1 d_2 = 49 - d_1^2 - d_2^2$. Square both sides again and see what you get.

 

[PsychonautQQ]

but d_1 and d_2 are |z (+/-) 1| not |z(+/-)i|

 

[Ray Vickson]

but d_1 and d_2 are |z (+/-) 1| not |z(+/-)i|

So, make the necessary changes. It really does not matter.

 

[PsychonautQQ]

I'm not sure if this is as neat as you make it look, I mean squaring the d_1^2 and d_2^2 is a messy ordeal in itself. I need to extract the real and complex parts from the d_1 and d_2 before i beat it into elliptical form, right?

 

[PsychonautQQ]

Squared it again and got 2401 - 98(d_1)^2 - 98(d_2)^2 - 2(d_1*d_2)^2 + d_1^4 + d_2^4 = 0

These d^4 are scaring me, this looks like it's just getting more messy, am I doing this right?

 

[Ray Vickson]

Squared it again and got 2401 - 98(d_1)^2 - 98(d_2)^2 - 2(d_1*d_2)^2 + d_1^4 + d_2^4 = 0

These d^4 are scaring me, this looks like it's just getting more messy, am I doing this right?

Expand it out in detail. You cannot predict what will happen if you do not try it. In other words, $d_1^2 = (x-1)^2+y^2 = x^2 - 2x + 1 + y^2$, and $d_1^4 = (x^2+y^2-2x+1)^2 = \cdots$, etc.

 

[haruspex]

if I square both sides I wind up with a big mess

Not sure that it is any easier, but you can limit the mess by working as far as possible with $z$ and $\overline z$. First, derive an expansion for |z-1|².

 

[Ray Vickson]

Squared it again and got 2401 - 98(d_1)^2 - 98(d_2)^2 - 2(d_1*d_2)^2 + d_1^4 + d_2^4 = 0

These d^4 are scaring me, this looks like it's just getting more messy, am I doing this right?

Actually, instead of what I suggested in #7, write $2 d_1 d_2 = 7^2 - d_1^2 - d_2^2$, and write out the right-hand side here as $A x^2 + B y^2 + Cx+D$ before squaring it. That makes things a lot easier----still messy, but easier.

 

[Ray Vickson]

I'm not sure if this is as neat as you make it look, I mean squaring the d_1^2 and d_2^2 is a messy ordeal in itself. I need to extract the real and complex parts from the d_1 and d_2 before i beat it into elliptical form, right?

You are on the wrong track, We have $|z - (a + bi)| = \sqrt{(x-a)^2+(y-b)^2}$, so it is not complicated at all.