Wednesday, 26 November 2014

complex numbers - Find all zinBbbC such that |z+1|+|z1|=4

I'd like to find all points of the complex plane which satisfy



|z+1|+|z1|=4.



I know this is an ellipsis with foci 1 and 1, and I know that the answer is




3x2+4y212,



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+(x1)2+y2(x+1)2+y2=8.



Pass x2+y2+1 to the RHS (right hand side), then




(x1)2+y2(x+1)2+y2=7x2y2.



Now, I would have to square both sides of the equation like this,



((x1)2+y2)((x+1)2+y2)=(7x2y2)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



(x1)2+y2(x+1)2+y2=(7x2y2)




in which case also satisfy (2) but not (1)!



So I would get an incorrect solution.

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