Let a, b be two complex numbers in the left half-plane. $\int f(z) \leq 1$ and M-T suggest \exists c$
The geometric view is that $|a-b|$ is the length of a curve between $e^{ia}$ and $e^{ib}$ - namely, the curve $e^{i\theta}$ for $\theta\in[a,b]$ (or $[b,a]$, depending on the order of the values.)

But the shortest distance between two points is the linear distance, which is for these points $\left|e^{ib}-e^{ia}\right|$.
Lemma: For $h&gt;0,1-\cos h &lt; h^2/2.$ Proof: Since $\sin t &lt; t$ for $t&gt;0,$ we have

$$1-\cos h = \int_0^h\sin t \, dt &lt; \int_0^h t \, dt = h^2/2.$$

Thm: $|e^{ih}-1| &lt; h$ for $h&gt;0.$ Proof: $|e^{ih}-1|^2 = 2(1-\cos h) &lt; h^2,$ where we have applied the lemma. Now take square roots.

Corollary: If $a&lt;b,$ then $|e^{ib} - e^{ia}| &lt; b-a.$ Proof:

$$|e^{ib} - e^{ia}| = |e^{ia}(e^{i(b-a)} - 1)| = |e^{i(b-a)} - 1| &lt; b-a,$$

the last inequality coming from the theorem.