Computing the maximum modulus For each $a\in \mathbb C$ define $f_a:\mathbb C\to \mathbb C$ by $f_a(z)=\exp(z)+a$. I am primarily interested in real values $a\in (-\infty,-1)$.
For each $r\in [0,\infty)$ define $M_a(r)=\max\{|f_a(z)|:|z|=r\}$.  So $M_a(r)=|z'|$ for some point $z'$ in the image of the circle of radius r.  Is there a more explicit (numerical) formula for $M_a(r)$ in terms of $a$ and $r$? 
 A: An answer before the numerics start: First, note that for $w=u+iv \in   \mathbb{C}$ and a fixed $a \in \mathbb{R}$ we have $|w+a|=\sqrt{u^2+v^2+2au+a^2}=\sqrt{|w|^2 + 2au+a^2}$ . Next, consider the image under the complex exponential of the circle $|z|=r>0$. Using polar coordinates, we get $|\exp (r\cos t+i r\sin t)|=e^{r\cos t}$. Thus to get $M_a(r)$ we need to maximize the function $g(t)=\sqrt{e^{2r\cos t}+2ae^{r\cos t}\cos (r\sin t)+a^2}$ for $t \in [0,2\pi]$.
 For $r=0$ we get $M_a(0)=|\exp 0 +a|=|1+a|$. 
A: A partial solution: As mentioned by @MargaretFriedland, the desired $M_a(r)$ is the absolute maximum of  $g(t):=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$. Notice that this is an even function, so it suffices to maximize over $[0,\pi]$. If $t\in\left[\frac{\pi}{2},\pi\right]$, then $\cos t<0$. So ${\rm{e}}^{2r\cos t}\in [0,1]$ and, keeping in mind that $a$ is assumed to belong to  $(-\infty,-1)$, we have:
$$\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}\leq\sqrt{1-2a+a^2}=1-a.$$ 
This value is achieved at $t=\pi$. Consequently: 
$M_a(r)=\max\left\{1-a,\max_{\,t\in \left[0,\frac{\pi}{2}\right]}g(t)\right\}$. One can elaborate more by conditioning on $r\geq 0$. 


*

*If $r\leq\pi$, as $t$ varies in $\left[0,\frac{\pi}{2}\right]$, $r\sin t$ is increasing and takes its values in the interval $[0,\pi]$ on which cosine is decreasing. We conclude that $g(t)=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$ is decreasing on $\left[0,\frac{\pi}{2}\right]$ as every non-constant term under the square root is. Hence in this case $$M_a(r)=\max\{1-a,g(0)=|{\rm{e}}^r+a|\}.$$

*If $r>\pi$, the absolute maximum $M_a(r)$ is larger than $1-a$. To see this, notice that there exists $t_0\in\left[0,\frac{\pi}{2}\right]$ with $\sin t_0=\frac{\pi}{r}$. We have:
$$g(t_0)=\sqrt{{\rm{e}}^{2r\cos t_0}-2a{\rm{e}}^{r\cos t_0}+a^2}={\rm{e}}^{r\cos t_0}-a={\rm{e}}^{r\sqrt{1-(\frac{\pi}{r})^2}}-a> 1-a.$$ Thus in this case
$$
M_a(r)=\max_{\,t\in \left[0,\frac{\pi}{2}\right]}g(t)\geq{\rm{e}}^{\sqrt{r^2-\pi^2}}-a.
$$
Added: It is also interesting to notice that always $M_r(a)\in\big[\,|{\rm{e}}^r+a|,{\rm{e}}^r-a\big)$; the function $g(t)$ can never attain 
${\rm{e}}^r-a$ because that requires to simultaneously have $\cos t=1$ and $\cos(r \sin t)=-1$. Thus the negativity of $a$ is really what makes this problem interesting. As a result: $\big|M_r(a)-({\rm{e}}^r-a)\big|<2a$.
Fixing $a$, the ratio $\frac{M_r(a)}{{\rm{e}}^r-a}$ tends to $1$ as $r\to\infty$. 
A: Assuming that a closed formula is not accessible, possibly maxmod, which calculates the maximum modulus of a complex polynomial on the unit disk, would speed a numerical attack. (Polynomial order of a few hundred is not a problem.)
