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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:

$\left|\frac{a+(b-1)i}{a+(b+1)i}\right|<1\\ \left|\frac{(a+(b-1)i)(a-(b+1)i)}{a^2+(b+1)^2}\right|<1\\ \left|\frac{a^2-(ab+a)i+(ab-a)i+(b^2-1)}{a^2+b^2+2b+1}\right|<1\\ \left|\frac{a^2+b^2-1-2ai}{a^2+b^2+2b+1}\right|<1\\ \left|\frac{a^2+b^2-1}{a^2+b^2+2b+1}-\frac{2ai}{a^2+b^2+2b+1}\right|<1\\ \left(\frac{a^2+b^2-1}{a^2+b^2+2b+1}\right)^2+\left(\frac{2ai}{a^2+b^2+2b+1}\right)^2<1\\ \frac{a^4+2a^2+2a^2b^2-2b^2+b^4+1}{a^4+2a^2+4a^2b+2a^2b^2+4b+6b^2+4b^3+b^4+1}<1$

ahorn
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2 Answers2

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Rewrite the inequality as $|z-i|<|z+i|$. Then think geometrically in terms of distances from the point $z$ to the points $\pm i$.

Hans Lundmark
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$|z-i|<|z+i| \implies |x+(y-1)i|<|x+(y+1)i| \implies x^2+y^2-2y+1<x^2+y^2+2y+1\implies y>0$

is just an algebraic way to show what Hans Lundmark said

ahorn
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Eli Elizirov
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  • sorry i will edit – Eli Elizirov Jun 24 '14 at 21:43
  • Sorry, you were right the first time. $y>0$. The distance to $i$ above the real line is shorter than the distance to $-i$ and $|z-i|$ represents the distance from $z$ to $i$.

    $-2y<2y \Rightarrow -y<y \Rightarrow -2y<0 \Rightarrow y>0$.

     – ahorn Jun 24 '14 at 22:05