1. ## Complex function help

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

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

Since $|z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0< 1-|z|^{2}$ and $|1-z|^{2}>0$ for all $z$, $f$ 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.

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

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

Since $|z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0< 1-|z|^{2}$ and $|1-z|^{2}>0$ for all $z$, $f$ 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.

Check carefully the denominator in the third equality of your "proof": do you think that $(1+z)(1+\overline{z})=1-|z|^2$?? I don't think so.
A similar mistake happened with the numerator in the second expression...

Tonio

3. what would be the correct approach? Should I write out z in the form x+iy and expand everyting out, distribute the i, and group?

4. You have made a mistake in the fractions: $\dfrac{(1-z)}{(1+z)}\dfrac{(1+\overline{z})}{(1+\overline{z} )}$.
Now $(1-z)(1+\overline{z})=1-2i\text{Im}(z)-|z|^2$.
And $(1+z)(1+\overline{z})=|1+z|^2$

So, $\displaystyle \text{Im}(i~f(z))=\dfrac{1-|z|^2}{|1+z|^2}$.

5. Originally Posted by Plato
You have made a mistake in the fractions: $\dfrac{(1-z)}{(1+z)}\dfrac{(1+\overline{z})}{(1+\overline{z} )}$.
Now $(1-z)(1+\overline{z})=1-2i\text{Im}(z)-|z|^2$.
And $(1+z)(1+\overline{z})=|1+z|^2$

So, $\displaystyle \text{Im}(i~f(z))=\dfrac{1-|z|^2}{|1+z|^2}$.
So, just to be clear, the conjugate of $(1+z)$ is $(1-\bar{z})$ and the conjugate of $(1-z)$ is $(1+\bar{z})$? If so, this means I alwars change the sign and take the conjugate of the complex number in the expression. Correct?

Now, since $|z|<1\Leftrightarrow\\|z|^{2}<1\Leftrightarrow\\0< 1-|z|^{2}$ and $|1+z|^{2}>0$ for all $z$ in the domain, it follows that $f$ maps fromt he unit disk into the upper half-plane.

Is this a sufficient answer to the question?

Thanks for the help

6. Originally Posted by Danneedshelp
So, just to be clear, the conjugate of [tex](1+z)[\math] is $(1-\bar{z})$ and the conjugate of $(1-z)$ is $(1+\bar{z})$?
That is completely wrong.
First. To rationalize a complex fraction multiply by the conjugate of the denominator.

So $\dfrac{z}{w}=\dfrac{z\overline{w}}{|w|^2}$

Moreover, $\overline{1+z}=1+\bar{z}$ AND $\overline{1-z}=1-\bar{z}$