Sketch the set of points determined by $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ Sketch the set of points determined by the following conditions $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ So I started by $(z-i)(\bar{z} -\bar{i}) + (z+i)(\bar{z} + \bar{i}) \le 4$
$z\bar{z} - z\bar{i} - i\bar{z} +  i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$
$=2z\bar{z} + i\bar{i}\le 4$
$=(x+iy)(x-iy) \le 2$
$=(x^2 + y^2)\le2$
$=x^2 + y^2 \le 2$ which gives me a circle with center at the origin with $r= \sqrt2$
But then I thought to do it a different way and then got confused on which way was correct. The second way is as follows. 
$\lvert z-i\rvert^2 + \lvert z+i\rvert^2 \le4$
$=\lvert a+(b-1)i\rvert^2 + \lvert a+(b+1)\rvert^2 \le 4$
$=a^2 + (b-1)^2 + a^2 + (b+1)^2 \le 4$
$=a^2 + b^2 -2b+1 + a^2 + b^2 + 2b+1 \le 4$
$=2a^2 +2b^2 \le 4$
$=a^2 + b^2 \le 1$ which gives me a circle with $r=1$
I guess my question is, which one is the right process?

While the algebraic route is simple, one actually can deduce the result from geometry alone. You have two points, separated by a distance of $c=2$, and you require that their distances $a,b$ to a point $z$ satisfy $a^2+b^2\leq c^2$. For the boundary points, for which this is an equality, we recognize the Pythagorean theorem and conclude that every point is the base point of some right triangle with hypotenuse $c=2$. But from Euclidean geometry we know that the base points of right triangles with a common hypotenuse all lie on a common circle with the hypotenuse as diameter. From such considerations we conclude that the boundary points are the unit circle and the region of interest is the unit disk.
