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**Danneedshelp** Q: Let $\displaystyle w=f(z)=i(\frac{1-z}{1+z})$. Show that $\displaystyle f$ maps the open unit disk $\displaystyle \{z\in{\mathbb{C}}:|z|<1\}$ into the upper half-plane $\displaystyle \{w\in{\mathbb{C}}:Im(w)>0\}$, and maps the unit circle $\displaystyle \{z\in{\mathbb{C}}:|z|=1\}$ to the real line.

Im am not sure how to approach this. I found a problem similar to this one in my text book. Here is what I have worked out so far:

$\displaystyle Im(f(z))=Im(i(\frac{1-z}{1+z}))=Im(i(\frac{|1-z|^{2}}{(1+z)(1-\bar{z})}))=Im(\frac{|1-z|^{2}i}{1-|z|^{2}})=\frac{|1-z|^{2}i}{1-|z|^{2}}$

Since $\displaystyle |z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0< 1-|z|^{2}$ and $\displaystyle |1-z|^{2}>0$ for all $\displaystyle z$, $\displaystyle f$ maps from the open disk of raduis 1 into the upper half plane.

I am not sure if I interpreting what I read correctly, so any help would be appreciated. Thanks.