Sum of two absolute values in complex plane I'm trying to find out all $z \in C$ that satisfy the following condition:
$|z+1|+|z-i|=3$ 
I understand that $|z|=r$ represents a circle with a radius of $r$.
I also understand that $|z+1|=r$ can be written as $|(x+1)+yi|=\sqrt{(x+1)^2+y^2}=r$ which can then be squared to get $(x+1)^2+y^2=r^2$ which represents the circle with a radius of r, with center in $(-1,0)$.
So, back to my problem: 
Unlike the example with $|z+1|=r$, where it is easy to square the equation, squaring 
$|z+1|+|z-i|=3$ written as $\sqrt{(x+1)^2+y^2}+\sqrt{x^2+(y-1)^2}=3$ equals:
$(x+1)^2+(y-1)^2+x^2+y^2+2\sqrt{(x+1)^2+y^2}\sqrt{x^2+(y-1)^2}=9$ which is a nightmare to solve, if at all possible.
I am sure there must be some elegant way to solve this kind of problem. The way I'm thinking is this:
$|z+1|$ by itself seems to represent a circle centered at $(-1,0)$, with an undefined radius, and $|z-i|$ seems to represent a circle centered at $(0,1)$, also with an undefined radius. However, the sum of those two radii must be 3. But I can't seem to wrap my mind around what this would represent (when drawn on Cartesian plane), or how to solve it analytically. So, how would one approach this problem? Even a hint would (probably) suffice.
 A: $|z+1|$ is the distance from the point $z$ to the point $-1$, and $|z-i|$ is the distance from $z$ to the point $i$. Thus, you’re looking at the set of all points $z$ such the sum of the distances from $z$ to $-1$ and $i$ is $3$. Which conic section is defined as the locus of points $P$ such that the sum of $|PA|$ and $|PB|$ is a constant for two fixed points $A$ and $B$?
A: Let $\sqrt{(x-a)^2+(y-b)^2}+\sqrt{(x-c)^2+(y-d)^2}=e$
$\sqrt{(x-a)^2+(y-b)^2}=e-\sqrt{(x-c)^2+(y-d)^2}$
Squaring we get, $(x-a)^2+(y-b)^2=e^2+(x-c)^2+(y-d)^2-2e\sqrt{(x-c)^2+(y-d)^2}$
or, $2e\sqrt{(x-c)^2+(y-d)^2}=e^2+(x-c)^2-(x-a)^2+(y-d)^2-(y-b)^2$
or, $2e\sqrt{(x-c)^2+(y-d)^2}=e^2+(2x-c-a)(a-c)+(2y-d-b)(b-d)$
or, $e\sqrt{(x-c)^2+(y-d)^2}=x(a-c)+y(b-d)+f$ where $2f=e^2-(c+a)(a-c)-(d+b)(b-d)$
Squaring we get, $e^2((x-c)^2+(y-d)^2)=(x(a-c)+y(b-d)+f)^2$
$\implies x^2(e^2-(a-c)^2)-2xy(a-c)(b-d)+y^2(e^2-(b-d)^2)-2x(e^2c+f(a-c))-2y(e^2d+f(b-d))+e^2(c^2+d^2)-f^2=0$
Using this, $4(a-c)^2(b-d)^2-4(e^2-(a-c)^2)(e^2-(b-d)^2)$
$=4e^2((a-c)^2+(b-d)^2-e^2)$
Now $e=\sqrt{(x-a)^2+(y-b)^2}+\sqrt{(x-c)^2+(y-d)^2}$
$=|(x,y)-(a,b)|+|(x,y)-(c,d)|<|(a,b)-(c,d)|$
So, $(a-c)^2+(b-d)^2-e^2<0$, so the locus of $(x,y)$ is an ellipse ,but not a circle unless the coefficient of $xy$ is $0$.

Alternatively,
$\sqrt{(x+1)^2+y^2}+\sqrt{x^2+(y-1)^2}=3$
$\sqrt{(x+1)^2+y^2}=3-\sqrt{x^2+(y-1)^2}$
Squaring we get, $ (x+1)^2+y^2=9+x^2+(y-1)^2-6\sqrt{x^2+(y-1)^2}$
$2x+2y-8=-6\sqrt{x^2+(y-1)^2}$
$x+y-4=-3\sqrt{x^2+(y-1)^2}$
Squaring we get, $(x+y-4)^2=9(x^2+(y-1)^2)$
$8x^2-2xy+8y^2+8x-10y-7=0$
Using this or this, $(-2)^2-4\cdot 8 \cdot 8<0$
So, the curve is an ellipse.
Using this , we can remove $xy$ terms.
