Given the curves:
- $x=\frac{\pi}{4}$
- $y=\frac{\pi}{4}$
Find the images by the transformation $w(z)=1+e^{-z}$
What I did
First decompose the tranformation function $w(z)$ into $w(z)=u(x,y)+iv(x,y)$. So,$$u(x,y)=1+e^{-x}\cos y$$ $$v(x,y)=-e^{-x}\sin y$$ Then, for the first curve $x=\frac{\pi}{4}$ it's possible to find the relation $$(u-1)^2+v^2=e^{-\frac{\pi}{2}}$$ This is a circle in the uv-plane.
For the other curve, find the relation $$v=-u+1$$ .
My question is if I define an orientation to the curves how the orientation changes when the transformation is applied?
If you parameterize the first curve as $$ \gamma(t) = \frac \pi 4 + it \, , \quad t \in \Bbb R $$ then $$ w(\gamma(z)) = 1 + e^{-\frac \pi 4 - it} = 1 + e^{-\frac \pi 4} (\cos(t) - i \sin(t)) \, . $$ Therefore if $t$ increases, $\gamma(t)$ moves “upwards” and $w(\gamma(z))$ rotates on the circle $|z-1| = e^{-\frac \pi 4}$ in a clockwise direction.
A similar approach can be used for the second curve.