Finding a simpler representation for the set $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$ 
Sketch the following set: $\{z\in\mathbb{C}|\|z+i\|+\|z-i\|=4\}$

I tried thinking about it geometrically, where we are interested in all triangles of the form (x,y-1), (x,y+1), (0,0) and we want the sum of the sides next to (0,0) be 2. Intuitively this makes me think the solution will be an ellipse of sort (where $\pm i$, and $\pm3$ being in the set helps this feeling), but I wasn't able to prove it.
Algebraically, I tried squaring both sides, to get 
$$ \left|z-i\right|^{2}+\left|z+i\right|^{2}+2\left|z-i\right|\left|z+i\right|=16 $$
and while the first two summands behave nicely 
$$\begin{aligned}\left|z-i\right|^{2} & = & \left(z-i\right)\overline{\left(z-i\right)}\\
 & = & z\overline z-i\overline z+iz+1\\
 & = & \left|z\right|^{2}+i\left(z-\overline{z}\right)+1\\
 & = & \left|z\right|^{2}+i\left(2i{\rm Im}z\right)+1\\
 & = & \left|z\right|^{2}-2{\rm Im}z+1
\end{aligned}$$
and $$\begin{aligned}\left|z+i\right|^{2} & = & \left(z+i\right)\overline{\left(z+i\right)}\\
 & = & z\overline z-iz+i\overline{z}+1\\
 & = & \left|z\right|^{2}-i\left(z-\overline z\right)+1\\
 & = & \left|z\right|^{2}+2{\rm Im}z+1
\end{aligned}$$
the third one leaves me a bit stuck. Any help on proceeding?

This is the "two nails and fixed cord" construction of an ellipse.
Computing the minor axis:
For $z = a + 0\cdot i$ we have
$$
\lVert z+i \rVert + \lVert z-i \rVert
= \sqrt{a^2 + 1^2} + \sqrt{a^2 + (-1)^2}
= 2 \sqrt{1 + a^2} = 4 \iff \\
1 + a^2 = 4 \iff \\
a = \sqrt{3}
$$
For $z = 0 + b\cdot i$ we have
$$
\lVert z+i \rVert + \lVert z-i \rVert
= \sqrt{0^2 + (b+1)^2} + \sqrt{0^2 + (b-1)^2}
= b + 1 + b - 1 \\
= 2b = 4 \iff \\
b = 2
$$

We have $a = \sqrt{3}$ and $b=2$, so it should be
$$
\left( \frac{x}{\sqrt{3}} \right)^2 + \left( \frac{y}{2} \right)^2 = 1
$$
Now lets try this from scratch:
$$
4 = \lVert z+i \rVert + \lVert z-i \rVert \\
= \sqrt{x^2 + (y+1)^2} + \sqrt{x^2 + (y-1)^2} \iff \\
= \sqrt{x^2 + y^2 + 1 + 2y} + \sqrt{x^2 + y^2 + 1 - 2y} \iff \\
16 = 2(x^2 + y^2 + 1)+2\sqrt{(x^2 + y^2 + 1)^2 - 4y^2} \iff \\
8 = x^2 + y^2 + 1 + \sqrt{(x^2 + y^2 + 1)^2 - 4y^2} \iff \\
(8 - (x^2 + y^2 + 1))^2 = (x^2 + y^2 + 1)^2 - 4y^2 \iff \\
64 + (x^2 + y^2 + 1)^2 - 16  (x^2 + y^2 + 1) = (x^2 + y^2 + 1)^2 - 4y^2 \iff \\
64 = 16  (x^2 + y^2 + 1) - 4y^2 = 16 x^2 + 12 y^2 + 16  \iff \\
48 = 16 x^2 + 12 y^2 \iff \\
12 = 4 x^2 + 3 y^2 \iff \\
1 =x^2 / 3 + y^2 / 4 \iff \\
1 = \left(\frac{x}{\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2
$$
