I am trying to use Rouche's Theorem somehow but I can't seem to be able to find a proper function to compare $f_a(z)$ with.
I tried $g(z)=z+a$ but then I can't deal with the $e^z$ term.
Any suggestions or solutions?
Edit:
Suppose $f(z)$ has a root in $Re(z)<0$. This implies that $|z+a|\leq |e^z|=|e^{Re(z)}|<1$. Hence, z has to lie in open ball around $-a$
For $z$ on the unit circle around $-a$
$|g(z)-f(z)|=|e^z|=e^{Re(z)}<1=|g(z)|$
By Rouche's this implies $f(z)$ has 1 zero.
Is this correct? But how does one show that the root is real.
You have only shown that $f$ has exactly one zero in the disk with center $-a$ and radius one. Instead of that disk consider arbitrary rectangles $$ R= R(x_1, x_2, y) = \{ z \mid x_1 < \operatorname{Re}(z) < x_2 , -y < \operatorname{Im}(z) < y \} $$ where $x_1 < -a - 1$,$ -a + 1 < x_2 < 0$ and $y > 1$. Then your estimate holds on the boundary of $R$: $$ |f(z) - (z+a)| = |e^z| = e^{\operatorname{Re}(z)} < 1 < |z + a| $$ so that $f$ and $(z+a)$ have the same number of zeros (i.e.: one) in $R$. Since the rectangles $R$ exhaust the left half-plane, it follows that $f$ has exactly one zero on the left half-plane.
Finally, a simple analysis of $f$ restricted to real arguments shows that $f(x) = 0$ for some negative $x \in \Bbb R$.