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 $r$? 



Computing the maximum modulus
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 $r\geq 0$. 


1. 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 $$M\_a(r)=\max\{1-a,g(0)=|{\rm{e}}^r+a|\}.$$
2. 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 $r\to\infty$. 


