How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\  \epsilon\  C : |Z-1|+|Z+1|=4\}$  
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=16$
$x^2+y^2=7$
$(x-0)^2+(y-0)^2=\sqrt7$
=This is a circle with center $0$ and radius $\sqrt7$  
My answer is different from the correct answer given: "This is an ellipse with foci at $-1$ and $1$ passing through $2$"
I have no idea how to get to this answer. Could someone please help me here? 

Hint:
For brevity, let us write $R_1:=|z-1|, R_2:=|z+1|$, which are square roots of a quadratic polynomial.
Squaring the sum,
$$(R_1+R_2)^2=R_1^2+2R_1R_2+R_2^2=a^2.$$
Then
$$4R_1^2R_2^2=(a^2-R_1^2-R_2^2)^2
=a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2+2R_1^2R_2^2,$$
$$a^4+R_1^4+R_2^4-2a^2R_1^2-2a^2R_2^2-2R_1^2R_2^2=0,$$
$$a^4+(R_1^2-R_2^2)^2-2a^2R_1^2-2a^2R_2^2=0.$$
As $R_1^2-R_2^2$ simplifies to the square of a linear expression, you get a quadratic equation, i.e. a conic.
