If $\displaystyle C$ is the frontier of the triangle with vertices $\displaystyle z_1=0$, $\displaystyle z_2=3i$ and $\displaystyle z_3=-4$, show that $\displaystyle \left| \int _C (e^z- \overline z)dz \right| \leq 60$ without calculating the integral.

My attempt : I drew the sketch of $\displaystyle C$.

Then I wrote $\displaystyle \left| \int _C (e^z- \overline z)dz \right| \leq \int _C \left| (e^z- \overline z) \right| dz$.

I also wrote that $\displaystyle \int _C dz = \sqrt{4^2+3^3}+4+3=12$, hence I'm done if I show that $\displaystyle |e^z- \overline z|\leq 5$ for $\displaystyle z \in C$. (Correct me please if I'm wrong saying that $\displaystyle z \in C$, where $\displaystyle C$ denotes the frontier of the triangle.)

Here's where I'm stuck. I thought about calculating the image of the interior of the triangle under the transformation $\displaystyle z \mapsto e^z$ but I believe this is a long work. Also it wouldn't finish there. I believe I'm missing something obvious.