Plot $|z - i| + |z + i| = 16$ on the complex plane 
Plot $|z - i| + |z + i| = 16$ on the complex plane

Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.
I get to the point:
$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$
$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$
It seems like I'm and doing something the hard way.
 A: He asked for a plot:

from here, where I used 
$
\sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2}
$
from the currently accepted answer.
A: Just keep going:
$$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$$
$$\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 127 - (x^2+y^2)$$
$$(x^2 + (y - 1)^2)(x^2 + (y + 1)^2) = (127 - (x^2+y^2))^2$$
$$x^4 + x^2[(y - 1)^2 + (y + 1)^2] + [(y+1)(y-1)]^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ (x^2+y^2)^2$$
$$x^4 + 2x^2y^2 +2x^2 + (y^2-1)^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$$
$$x^4 + 2x^2y^2 +2x^2 + y^4-2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$$
$$2x^2 -2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)$$
$$x^2(2+2\cdot127) +y^2(-2+2\cdot127) = 127^2 -1$$
$$256x^2 +252y^2 = 16128$$
$$\frac {x^2} {63} +\frac {y^2} {64} = 1$$
A: Don't we all love algebraic solutions...
The ellipse has an equation of the form
$$(\frac {x} {a})^2 + (\frac {y} {b})^2 = 1$$
The focals are aligned on the y axis, therefore


*

*a is the side of a rectangle triangle, the other site being 1 and the hypothenuse 16/2:


$$a = \sqrt{8^2-1}$$


*

*b is 16/2.


The equation of the ellipse is
$$(\frac {x} {\sqrt{63}})^2 + (\frac {y} {8})^2 = 1$$
Or, to write it as above
$$\frac {x^2} {63} + \frac {y^2} {64} = 1$$
A: Maybe it's quicker way.
Equation $|z-i| + |z+i| = 16$ is equivalent to
$$
\sqrt{x^2 + (y-1)^2} = 16 - \sqrt{x^2 + (y+1)^2}.
$$
Squaring both sides you obtain
$$
x^2 + (y-1)^2 = 256 - 32\sqrt{x^2+(y+1)^2} + x^2 + (y+1)^2.
$$
Some terms cancel out hence you get
$$
8\sqrt{x^2 + (y+1)^2} = 64 + y,
$$
and
$$
64(x^2 + (y+1)^2) = 64^2 + 128y + y^2.
$$
Finally
$$
64x^2 + 63y^2 = 64 \cdot 63.
$$
Now it is easy to write ellipse equation
$$
\frac{x^2}{63} + \frac{y^2}{64} = 1.
$$
A: I think you are on the right track.
Wolframalpha also doesn't have a "simpler form" of the algebraic equation.
(see http://www.wolframalpha.com/input/?i=%28%7Cx%2Biy-i%7C%2B%7Cx%2Biy%2Bi%7C%29%5E2%3D256)
Maybe this question is meant to be done on computer. Usually questions done by hand will ask you to "sketch" the graph.
A: Sketching the graph gives a centrally symmetric ellipse with the imaginary axis as the major axis. The length of the half the major axis is clearly 8 (square 64), and pythagoras gives the square of half the minor axis as 63. This is reflected in the answer obtained by algebraic manipulation.
$$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$$
$$\sqrt{(x^2 + y^2 + 1) - 2y}\sqrt{(x^2 + y^2 + 1) + 2y} = 127 - x^2 - y^2$$
Square both sides:
$$((x^2 + y^2 + 1) - 2y)((x^2 + y^2 + 1) + 2y) = (127 - x^2 - y^2)^2$$
$$(x^2+y^2+1)^2-4y^2 = 127^2+x^4+y^4-254x^2-254y^2+2x^2y^2$$
Cancelling as we go and gathering terms in the obvious way:
$$256x^2+252y^2=127^2-1=128 \times 126$$
(difference of two squares)
The divisions now become easy and we reach the canonical form:
$$\frac {x^2} {63} + \frac {y^2} {64} = 1$$
