## henpen 3 years ago A complex question

1. henpen

$\left| z-1 \right|+\left| z+1 \right|=4$ Which turns into the messy $\sqrt{(a-1)^2+b^2}+\sqrt{(a+1)^2+b^2}=4$ Which I have no idea about how to solve graphically.

2. henpen

(Where z=a+ib)

3. henpen

(And by 'graphically' I mean plotting the possible values of z on the complex plane)

4. henpen

http://www.wolframalpha.com/input/?i=sqrt%28%28x%2B1%29%5E2%2By%5E2%29%2Bsqrt%28%28x-1%29%5E2%2By%5E2%29%3D4 Can anyone explain why the graph is an ellipse?

OK

| a + bi - 1| + |a + bi + 1| = 4

| (a -1) + (b)i | + | (a+1) + (b)i | = 4

Now let me find the modulus which is given by sqrt( re(z)^2 + Im(z)^2)

$\sqrt{(a +1)^2 + b^2} + \sqrt{(a-1)^2 + b^2} = 4$

Then you should square both sides

Actually that will be hard

So move one of the square roots on the left then square both sides

$(\sqrt{(a+1)^2 + b^2} )^2= (4 - \sqrt{(a-1)^2 + b^2})^2$

Which will give you this

15. henpen

There's still a horrible square root sign there, though.

$(a+1)^2 + b^2= 16 - 8\sqrt{(a-1)^2 + b^2} + (a-1)^2 + b^2$

I will get rid of it soon

Ok now i am going to cancel some stuff and simplify this equation

$a^2 + 2a + 1 + b^2 = 16 - 8\sqrt{(a-1)^2 + b^2} + a^2 - 2a + 1 + b^2$

Now i can simplify further

$4a -16 = -8\sqrt{(a-1)^2+b^2}$

22. henpen

4a-16=-8sqrt(a^2-2a+1+b^2)

Now I am going to square both sides again

$(4a -16)^2 = 64((a-1)^2 + b^2)$

Finally we have no more square roots

Now i am going to expand

$16a^2 -128a + 256 = 64 a^2 -128a + 64 + 64b^2)$

28. henpen

it ends up as 192=48a^2+64b^2 I think

Now simlify...

SSo you get

31. henpen

That actually agrees with the wolfram link I posted earlier. Thanks very much- I had little idea how to deal with square roots efficiently before!

Final step: complete the square on a term

oops

Sorry yeah that was the answer 192 = 48a^2 +64b^2

And thats an eclipse.