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? 
 A: Or in another but similar parametrization to Yves':
\begin{align}
\sqrt{a+b}+\sqrt{a-b}&=4\\
\text{after squaring: }a+\sqrt{a^2-b^2}&=8\\
\text{rearrange and square: }a^2-b^2&=(8-a)^2=64-16a+a^2\\
16a-b^2&=64
\end{align}
where $a=x^2+y^2+1$ and $b=2x$.
A: Hint:
The equations
$$\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$$
$${{(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$$
are not equivalent.
Squaring the first one yields
\begin{align}
\left[\sqrt{{(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}\right]^2&=4^2\\
(x-1)^2+y^2+2\sqrt{{(x-1)}^2 + y^2}\sqrt{{(x+1)}^2+y^2}+(x+1)^2+y^2&=16
\end{align}
A: 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.
A: That's a good question. Unfortunately you can't just square term by term like that. When you write this out, let z=x+iy. If you do that you will get a very complicated algebraic term. 
((x-1)^2 + 2*y^2 + (x+1)^2)^2 = 16
As LutzL had brilliantly stated you can make a substitution, in this case a make a parametrization, where a = (x-1)^2 and b = (x+1)^2
However there is still a much easier solution. If you have a textbook called Advanced Mathematics by Terry Lee, he goes through how this can be done. This is an ellipse. The 4 at the end of the equation, instead of representing the radius of a circle, this represents the major axis of an ellipse.
We know for an ellipse x = 2a (this is equation of major axis).
Hence 2a = 4              Therefore a = 2
We now know our semi-major axis is 2. If we use form of an ellipse where 
(x^2)/(a^2) + (y^2)/(b^2) = 1 
Hence we can write (x^2)/4 + (y^2)/(b^2) = 1
The foci represents the left part of the equation in this case going from z-1 to z+1. Set these to equal 0 and hence the foci are (1,0) and (-1,0).
Now let S = focus
The equation of the focus is given by: S = ae
We substitute our data: S=1, a=2(semi-major axis)
e=1/2 (eccentricity)
From this we can then find the semi-minor axis.
We know the formula b^2 = a^2(1-e^2)
If we solve, we get b= sqrt(3)
Hence the equation of the ellipse is given by: x^2/4 + y^2/3 = 1
If you want you can now find directrices: Let D = Directrix
By definition D = +- a/e = +- 2/0.5 = +- 4
You can now graph this.
