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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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  • 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? – Theo Bendit Sep 14 at 11:55
  • 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. – limeeattack Sep 14 at 12:08
up vote 1 down vote accepted

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

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