Describe the set of points on the complex plane for which $|z-2| + |z+2|=4$... So, I know you can solve this instantly, just by using definition, but I want to do it the long way..
So, $$|x- i*y -2|+ |x-i*y +2| = 4$$ $$(x-2)^2 + y^2 + (x+2)^2 +y^2 =16$$ $$ x^2-4*x+4+y^2+x^2+4*x+ 4 +y^2=16$$
now, $x^2$ and $y^2$ were supposed to eliminate but they don't ...where am I wrong?
Of course the following is the "wrong" way to do the problem. The geometry, or the Triangle Inequality, tell us that since the distance between $-2$ and $2$ is $4$, our locus consists of all points in the line segment $[-2,2]$.
But we will calculate. If we go to coordinates, we get $$\sqrt{(x+2)^2+y^2}+\sqrt{(x-2)^2+y^2}=4.$$ Rewrite as $\sqrt{(x+2)^2+y^2}=4-\sqrt{(x-2)^2+y^2}$ and square both sides. There is very nice cancellation, and we get $$8x-16=-8\sqrt{(x-2)^2+y^2},$$ Divide through by $8$, and square both sides again. We get $$(x-2)^2=(x-2)^2+y^2.$$ Thus $y=0$.
So all candidates must be on the $x$-axis. We have done some squaring, and extraneous roots may have been introduced. So go back to the original equation, and put $y=0$. We get $|x+2|+|x-2|=4$. It is easy to verify that this is true precisely if $-2\le x\le 2$.