I'd like to find all points of the complex plane which satisfy
|z+1|+|z−1|=4.
I know this is an ellipsis with foci 1 and −1, and I know that the answer is
3x2+4y2≤12,
but I can't find a correct way of getting there.
First, I write z as x+iy and square both sides of the equation, then divide by 2 and get
x2+y2+1+√(x−1)2+y2√(x+1)2+y2=8.
Pass x2+y2+1 to the RHS (right hand side), then
√(x−1)2+y2√(x+1)2+y2=7−x2−y2.
Now, I would have to square both sides of the equation like this,
((x−1)2+y2)((x+1)2+y2)=(7−x2−y2)2,
but the problem is that I cannot assure that the RHS is not negative, so there could be a value for z such that (2) is satisfied but not (1), i.e it could exist z=x+iy which satisfies
√(x−1)2+y2√(x+1)2+y2=−(7−x2−y2)
in which case also satisfy (2) but not (1)!
So I would get an incorrect solution.
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