Q:

Strange way to define an ellipse in the complex plane

I'm asked to prove that 
$|z-i||z+i|=2$
defines an ellipse in the plane.
I have tried replacing $z = x+iy $  in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.
Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says. 
In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.

A:

This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.
In order to prove that it is not an ellipse, note that four of the points of this set are $\pm1$ and $\pm\sqrt3\,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$\left\{x+yi\in\mathbb{C}\,\middle|\,x^2+\frac{y^2}3=1\right\}.$$One of the points of this ellipse is $z_0=\frac12+\frac{3i}2$. But $|z_0-i|.|z_0+i|=\frac{\sqrt{13}}2$. Therefore, your set is not an ellipse.

A:

The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.

