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Given the function $f:\mathbb{C}\setminus\{-i\}\rightarrow \mathbb{C}\setminus \{1\}$, defined by $f(z)=\frac{z-i}{z+i}$.

I'm supposed to find the image for $f(\{z\mid\Im (z) > 0\})$. However I'm fairly uncertain on how to do this.

I know that $z = \frac{-(a+1)i}{a-1}$ for some $a\in \mathbb{C}\setminus \{1\}$, and i've tried writing z on the form $n+im$ for $n,m\in\mathbb{R}$, but that doesn't help me.

Any suggestions on how to solve it?

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  • $\begingroup$ It is a Mobius transform. It should map a half-space to either a half-space, ball, or complement of a ball. Maybe start by mapping three points on the boundary, and finding the unique circle or line that passes through their images? $\endgroup$ – Theo Bendit Sep 14 '18 at 11:55
  • $\begingroup$ The course just started so we haven't learned about Mobius transforms yet. I'll look it up and see if it's useful, thank you for the advice. $\endgroup$ – limeeattack Sep 14 '18 at 12:08
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Hint: With $z=x+iy$ simplify $$u+iv=\dfrac{z-i}{z+i}$$ where $w=u+iv$ is in the range of map. Then find $$1-u^2-v^2=((1-u)^2+v^2)y>0$$ which gives the result.

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  • $\begingroup$ I'm not quite sure what you mean, how does the last relation make sense? $\endgroup$ – limeeattack Sep 14 '18 at 18:41
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    $\begingroup$ means $u^2+v^2<1$. $\endgroup$ – Nosrati Sep 14 '18 at 18:49
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    $\begingroup$ note that $(1-u)^2+v^2$ and $y$ are positive. $\endgroup$ – Nosrati Sep 14 '18 at 18:50
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    $\begingroup$ simplify $(u+iv)(x+iy+i)=x+iy-i$ and consider real part and imaginary of both sides. $\endgroup$ – Nosrati Sep 14 '18 at 18:57
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    $\begingroup$ of course, drop $x$ between them. $\endgroup$ – Nosrati Sep 14 '18 at 19:19

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