Complex Distances are not so Complicated

The center of the ellipse is the midpoint of the two foci, which in this case is the origin; also, the two foci lie on the y-axis, so the ellipse will have a vertical major axis.

Then the equation of the ellipse will be $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ , with $b > a$ .

Let $2c$ be the distance between the two foci. Then for an ellipse with a vertical major axis, it is always true for any point on the ellipse that the sum of the distances to the two focal points is $2b$ , and also that $b^2 = a^2 + c^2$ . Thus $2b = 4 \implies b = 2$ , and then $c = 1 \implies 2^2 = a^2 + 1^2 \implies a = \sqrt{3}$ .

Thus the equation of the ellipse is $\dfrac{x^2}{3} + \dfrac{y^2}{4} = 1$ , i.e. $4x^2 + 3y^2 = 12$ , and our answer is $4 + 3 + 12 = \boxed{19}$

The equation $|z-i|+|z+i| = 4$ means that the sum of distances from point $z (x,y)$ to the two points $(0,-1)$ and $(0,1)$ is equal to 4, a constant. The means that the locus of $z(x,y)$ is an ellipse with points $(0,-1)$ and $(0,1)$ as the foci. The standard equation of an ellipse is $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ . Putting $x=0$ , we find the semiminor axis $a = \sqrt {2^2-1^2} = \sqrt 3$ . Putting $y=0$ , the semimajor axis $b = 2$ . Therefore, we have:

$\begin{aligned} \frac {x^2}{a^2} + \frac {y^2}{b^2} & = 1 \\ \frac {x^2}{3} + \frac {y^2}{4} & = 1 & \small \color{#3D99F6} \text{Multiply both sides by }12 \\ 4x^2+3y^2 & = 12 \end{aligned}$ .

Therefore, $A+B+C = 4+3+12 = \boxed{19}$ .

Relevant wiki: Ellipse

The sum of the distance from point $P(a,b)$ to $F_1(0,1)$ and $F_2(0,-1)$ is 4.So the ellipse is

$\frac{x^2}{2^2-1}+\frac{y^2}{(\frac42)^2}=1,4x^2+3y^2=12$

$|x+iy-i| + |x+iy+i| =4 \\ \sqrt{x^2+(y-1)^2} + \sqrt{x^2+(y+1)^2} = 4\\ x^2+y^2+1 + \sqrt{x^2+(y-1)^2} \sqrt{x^2+(y+1)^2} = 8 \\ \Big[x^2+(y-1)^2 \Big]\Big[x^2+(y+1)^2 \Big]=\Big[ 7-(x^2+y^2\Big]^2\\ x^4 +x^2(y+1)^2+ x^2(y-1)^2 +(y-1)^2(y+1)^2 = 49 - 14(x^2+y^2) + x^4+2x^2y^2+y^4 \\ 2x^2y^2+2x^2-2y^2 + 1 = 49 - 14(x^2+y^2) +2x^2y^2\\ 16x^2 +12y^2 = 48 \\ 4x^2 + 3y^2 =12 \\ \therefore A+B+C = \boxed{19}$