Finding it difficult to find a conformal mapping from the set
$\mathbb{C}\backslash$ $\{z :|Im(z)| \leq -Re(z)\}$
to the upper half plane.
Any advice will be very helpful
I know I can use $f(z) = e^z$ to map the strip $\{z : 0 < Im(z) < \pi\}$ to the upper half plane.
The region you're describing, when plotted (using Desmos, for example) is
Let's name it $G$.
The angle between the positive real axis and either edge is $3\pi/4$. We can work with the principal branch of the logarithm in $G$ so that we can define $f(z)=z^{4/3}$ in this region through the principal branch of the logarithm (of course).
What I mean here is that $$ f(z)=\exp\left(\frac{4}{3}\log(z)\right) $$ where $\log$ denotes the branch of the logarithm defined on the region $\mathbb{C}\setminus\{z\in\mathbb{C}\colon\: \mathrm{Re}(z)\leq0\text{ and }\mathrm{Im}(z)=0\}=G_1$.
This function maps $G$ conformally onto $G_1$. We have now transformed $G$ to $G_1$ conformally. Let's now transform $G_1$ to the right half plane $G_2=\{z\in\mathbb{C}\colon\:\mathrm{Re}(z)\geq0\}$ conformally. The function $g(z)=\sqrt{z}$ does the job.
Now transform $G_2$ to the upper half plane through a rotation and we are done. The conformal mapping we are looking for is given by the compositions of these maps: $$ e^{i\pi/2}(g\circ f)(z)=e^{i\pi/2}\sqrt{z^{4/3}} $$