Factor out into ellipse complex numbers Factor into equation of ellipse the following sets:
$$|z+2i|+ |z-2i| =6$$
$$|z-3|-|z+3|=4$$
I got to the part of taking one of the square roots and bringing it to the other side and then squaring both sides but can't seem to simplify it after that.
Detailed steps would be kindly appreciated, thanks!
 A: $$|x+yi+2i|+|x+yi-2i|=6$$
$$\sqrt{(x+yi+2i)(x-yi-2i)}+\sqrt{(x+yi-2i)(x-yi+2i)}=6$$
$$\sqrt{x^2+y^2+4y+4}+\sqrt{x^2+y^2-4y+4}=6$$
Square both sides:
$$8+2x^2+2y^2+2\sqrt{x^2+y^2+4y+4}\sqrt{x^2+y^2-4y+4}=36$$
Move $8+2x^2+2y^2$ to the RHS and square again:
$$4(16 + 8 x^2 + x^4 - 8 y^2 + 2 x^2 y^2 + y^4)=784 - 112 x^2 + 4 x^4 - 112 y^2 + 8 x^2 y^2 + 4 y^4$$
Leading to:
$$9x^2+5y^2=45$$
Or:
$$\frac{x^2}{5}+\frac{y^2}{9}=1$$
The second problem is similar. 
If you wanted to "cheat" or check your work, you could use the fact that in an ellipse, $r_1 + r_2 = 2a$, in your case $r_1 + r_2 = 6 \to a = 3$, and also the distance between focii is $4 = 2c$. Thus you could have found the coefficient of $y^2$ as $1/a^2=1/9$, and that of $x^2$ as $1/(a^2-c^2)=1/5$
A: 
Move 8+2x2+2y2 to the RHS and square again:

What if the resulting RHS is equal to -LHS for some x,y ?
When both sides squared, x,y would be a solution for this equation but not for the original one.
I mean: you can't square if not sure BOTH sides are not negative.
