<s>[INST] 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?
 [/INST] In your first derivation, you have a sign error:
$$z\bar{z} - z\bar{i} - i\bar{z} \oplus i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$$
The plus sign is circled for clarity. From there on, it works.
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