Can anyone help me with a complex number problem please?
Find the Cartesian equation for the locus of points $P(x, y)$ if $z = x + iy$ and: $|z+3|+|z-3|=8$. I tried four different situations such as when $$(z+3) \geq 0 \qquad\text{and}\qquad (z-3) \geq 0,$$ when $$(z+3) \leq 0 \qquad\text{and}\qquad (z-3) \leq 0,$$ when $$(z+3) \geq 0 \qquad\text{and}\qquad (z-3) \leq 0,$$ and when $$(z+3) \leq 0 \qquad\text{and}\qquad (z-3) \geq 0.$$ I got $x=4 $ and $x=-4$. I do not know how to proceed at this point and the answer given to this question is $7x^2+16y^2=112$.
In the complex numbers $|z-3|$ doesn't split into the cases $z-3\ge0$ and $z-3<0$. There's no order relation on the complex numbers (compatible with the operations).
Hint: the sum of the distances with two fixed points is constant. Does it ring a bell?
Otherwise, write the equation as $|z-3|=8-|z+3|$ and square, recalling that $|w|^2=w\bar{w}$ (the overline meaning conjugation).
Note that $|z-3|^2=(z-3)(\bar{z}-3)=z\bar{z}-3z-3\bar{z}+9$ and similarly for $|z+3|^2$. Observe also that, for $z=x+yi$, $z+\bar{z}=2x$. It should be easier, now.