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!

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The second one will be a hyperbola. –  coffeemath Feb 6 '13 at 22:58

$$|x+yi+2i|+|x+yi-2i|=6$$ $$\sqrt{(x+yi+2i)(x-yi-2i)}+\sqrt{(x+yi-2i)(x-yi+2i)}=6$$ $$\sqrt{x^2+y^2+4y+4}+\sqrt{x^2+y^2-4y+4}=6$$ 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 $1/(a^2-c^2)=1/5$

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Thanks for taking the time to make detailed steps ! –  DJ_ Feb 6 '13 at 23:52