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A complex question

Mathematics
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\[\left| z-1 \right|+\left| z+1 \right|=4\] 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.
(Where z=a+ib)
(And by 'graphically' I mean plotting the possible values of z on the complex plane)

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Other answers:

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
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
\[4a -16 = -8\sqrt{(a-1)^2+b^2}\]
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
\[16a^2 -128a + 256 = 64 a^2 -128a + 64 + 64b^2)\]
it ends up as 192=48a^2+64b^2 I think
Now simlify...
SSo you get
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.
No problem. =)

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