<s>[INST] Sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1 \}$ My question is to sketch the set $\{ z \in \mathbb{C} | \left|\frac{z-i}{z+i}\right|<1\}$ in the complex plane. 
I substituted $z$ for $a+bi$, but did not get anywhere:
$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\
\left|\frac{(a+(b-1)i)(a-(b+1)i)}{a^2+(b+1)^2}\right|<1\\
\left|\frac{a^2-(ab+a)i+(ab-a)i+(b^2-1)}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1-2ai}{a^2+b^2+2b+1}\right|<1\\
\left|\frac{a^2+b^2-1}{a^2+b^2+2b+1}-\frac{2ai}{a^2+b^2+2b+1}\right|<1\\
\left(\frac{a^2+b^2-1}{a^2+b^2+2b+1}\right)^2+\left(\frac{2ai}{a^2+b^2+2b+1}\right)^2<1\\
\frac{a^4+2a^2+2a^2b^2-2b^2+b^4+1}{a^4+2a^2+4a^2b+2a^2b^2+4b+6b^2+4b^3+b^4+1}<1$
 [/INST] $|z-i|<|z+i| \implies |x+(y-1)i|<|x+(y+1)i| \implies x^2+y^2-2y+1<x^2+y^2+2y+1\implies y>0$ 
is just an algebraic way to show what Hans Lundmark said
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