If $C$ is the frontier of the triangle with vertices $z_1=0$, $z_2=3i$ and $z_3=-4$, show that $\left| \int _C (e^z- \overline z)dz \right| \leq 60$ without calculating the integral.
My attempt : I drew the sketch of $C$.
Then I wrote $\left| \int _C (e^z- \overline z)dz \right| \leq \int _C \left| (e^z- \overline z) \right| dz$.
I also wrote that $\int _C dz = \sqrt{4^2+3^3}+4+3=12$, hence I'm done if I show that $|e^z- \overline z|\leq 5$ for $z \in C$. (Correct me please if I'm wrong saying that $z \in C$, where $C$ 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 $z \mapsto e^z$ but I believe this is a long work. Also it wouldn't finish there. I believe I'm missing something obvious.
2. Everything is fine with your idea, arbolis. Now, in order to prove that $|e^{z}-\overline z|$ is less than or equal to 5 over C, all you have to do is apply the triangle inequality. You get $|e^{z}-\overline z| \leq |e^{z}|+|z|.$ Since $|e^{z}|$ is equal to $e^{\mathrm{Re}(z)}$ and $e^{\mathrm{Re}(z)} \leq 1$ over C, you conclude that $|e^{z}-\overline z| \leq 1 + |z| \leq 1 + 4 =5.$ Done.