Find the image of the following region:
under the action of
$\displaystyle f(z)=e^{\bar{z}}$
(e raised to z conjugate)
I am lost on this one. Thanks for any advice with this one.
Let $\displaystyle \psi$ be the angle $\displaystyle y=\tfrac{x}{2}$ makes and $\displaystyle \phi$ be the angle $\displaystyle y=3x$ makes.
Your region can be described as, $\displaystyle \{ re^{i\theta} | r \in \mathbb{R}^{\times}, \psi <\theta < \phi \}$.
This means, $\displaystyle f(re^{i\theta}) = \exp ( \overline{re^{i\theta}} ) = \exp ( r e^{-i\theta}) = e^{r\cos\theta}e^{-ir\sin \theta}$.
Now $\displaystyle r\cos \theta,r\sin\theta$ ranges through all values of $\displaystyle \mathbb{R}^{\times}$. Therefore, the range is $\displaystyle r'e^{i\theta'}$ where $\displaystyle \theta'\in \mathbb{R}^{\times}$ and $\displaystyle r'>0,r'\not = 1$.