If an asymptotic expansion is valid as $\text{Re}(z)\to\infty$, must it be valid as $\text{Im}(z)\to\infty$? Suppose $f:U\to\mathbb C$ is analytic, where $U=\{z\in\mathbb C\,|\,\text{Re}(z)>0\}$ is the right half-plane. Suppose further that we have an asymptotic expansion (possibly divergent)
$$f(z)\sim\sum_{n=0}^\infty \frac{a_n}{z^n},\qquad\text{Re}(z)\to\infty.$$
Must this asymptotic series also be valid as $\text{Im}(z)\to\infty$ in the right half-plane? That is, must we have
$$f(x+iy)\sim\sum_{n=0}^\infty \frac{a_n}{(x+iy)^n},\qquad y\to\infty,$$
where $x,y\in\mathbb R$ and $x>0$? If not, under what conditions must this be true?

As a side note, I came across this issue for the trigamma function $\psi^{(1)}(z)$. In Gamelin's Complex Analysis, he proves that an asymptotic formula for this function holds for $\text{Re}(z)\to\infty$ (section XIV.2). Meanwhile, the same formula is listed in Abramowitz and Stegun's Handbook of Mathematical Functions claiming it is valid for $z\to\infty$ in $|\text{arg}(z)|<\pi$. So apparently my question (in fact, a stronger version) is true in this particular case. I just want to know why.
 A: Your particular result is true because it happens to be a  Laplace Transform of a function that does not grow too quickly near the postiive real line. 
Indeed, your book of Gamelin (page 367) expresses the trigamma $\psi'$  in this form,
$$ \psi'(z)=\frac{d}{dz} \frac{\Gamma'(z)}{\Gamma(z)} = \int_0^\infty e^{-zt}\frac{t}{1-e^{-t}} dt, \quad \Re z> 0.$$
This works because $$f(t) = \frac{t}{1-e^{-t}}$$ is exponential type of order $0+$ on a small neighbourhood of the positive real line, which fits into a more general result:

Theorem Let $\delta>0$, $\rho,\rho'>0$, with $\rho'>\rho$. 
  Suppose $U_\delta$ is a $\delta$-neighbourhood of $\mathbb R_+=(0,\infty)$ in $\mathbb C$, i.e. $$ U_\delta = \{t\in\mathbb C : d(t,\mathbb R_+) < \delta \}.$$ If $f$ is analytic of exponential order $\rho$ in $U_\delta$, then
  $$ \mathcal Lf(z) = \int_0^\infty e^{-zt}f(t) dt \simeq \sum_{k=0}^\infty \frac{f^{(k)}(0)}{z^{k+1}} $$
  as $z\to \infty$ in the set $\{\Re z > \rho'\}$.

Implicit above is the fact that $\int_0^\infty e^{-zt}f(t) dt$ converges for analytic functions of exponential order $\rho$ in $U_\delta$. We say $f$ is of exponential order $\rho>0$ in $D$ if there exists $M$ such that for every $t\in D$,
$$ |f(t)| \le M e^{\rho |t|}.$$
An example: if $f(t)=1$ then we can take any $\rho>0$ and $\mathcal Lf = \frac{1}{z}\simeq \frac1z + 0 + \dots $ is indeed valid on every half plane $ \{\Re z > \rho' \}$.
The proof is relatively straightforward; here's the sketch. The appearance of  derivatives strongly suggests integration by parts. Under the given assumptions, you can check that $f^{(k)}$ is also of exponential type $\rho$ for every $k$, which justifies the integration by parts. We record the constant that verifies this as $M_k$, i.e. 
$$ |f^{(k)}(z)| \le M_k e^{\rho |z|}.$$
Performing the integration by parts inductively leads to
$$\mathcal{L} f(z)=\sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{z^{k+1}}+\frac{1}{z^n} \int_{0}^{\infty} e^{-z t} f^{(n)}(t) d t$$
So we need to bound this last integral. One last integration by parts gives
\begin{align}
\left| \int_{0}^{\infty} e^{-z t} f^{(n)}(t) d t\right|
& \le \frac{|f^{(n)}(0)|}{|z|}  + \frac1{|z|}\int_0^\infty \underbrace{|e^{-tz} f^{(n+1)} (t)|}_{\Large\le M_{n+1} e^{-t \Re z+\rho t}}  dt
\\
&\le \frac{M_n}{|z|} + \frac{M_{n+1}}{|z|(\Re z-\rho)} 
\\
&\le  \frac{M_n}{|z|} + \frac{M_{n+1}}{|z|(\rho'-\rho)} 
\\&=: \frac{C_{n+1}}{|z|}
\end{align}
where $\rho<\rho'<\Re z$ was crucially used. This gives the required estimate to satisfy the definition of an asymptotic series (with asymptotic sequence $(z^{-k-1})_{k\ge 0}$).
Here is the bibliography from the description of my old course on Asymptotic Analysis, may be useful...

Books:
  We will not follow any particular book, but most of the material can be found in:
  
  
*
  
*C.F. Carrier, M. Krook and C.E. Pearson, Functions of a Complex Variable: theory and technique, Hodbooks.
  
*N.G. De Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing co. (3d ed.) (1970).
  
*P.P.G. Dyke, An Introduction to Laplace Transforms and Fourier Series, Springer Undergraduate Mathematics Series (2000).
  
*G. Hardy, Divergent Series, Clarendon Press, 1963/American Mathematical Society, 2000.
  
*R.B. Dingle, Asymptotic Expansions: Their Derivation and Interpretation, Academic Press (1973).
  

A: No. Indeed, the Riemann zeta function satisfies $\zeta(s) \sim 1$ as $\Re(x)\to\infty$, which can be viewed as a very simple asymptotic expansion; but its behavior on vertical lines is much more complicated.
(If you want an example with a more explicit asymptotic series, just add $\zeta(s)$ to a function of that type.)
A: How about $f(z) = e^{-z}$ 
We have $|e^{-z}| \to 0$ faster than any power of $1/z$ as $\mathrm{Re}\;z \to \infty$.  (So the asymptotic expansion in your sense has all coefficients $a_n = 0$.)
But of course $e^{-z}$ does not go to zero at all as $z\to \infty$ along the line $\mathrm{Re}\; z = 1$.
A: If it holds with $\psi'(z)$ I think it is because it works for meromorphic functions of the form $$f(z)=  z^{-m} g(z)+\sum_{k=1}^K\sum_n \frac{c_{n,k}}{(z+a_n)^{k}}$$ with $g$ a polynomial and $a_n\ge r>0$ and $\sum_{k=1}^K\sum_n \frac{|c_{n,k}|}{|a_n|^{k}}$ converges.
From the Laurent series of each term at $\infty$ we obtain for any fixed $J$ $$f(z)=  z^{-m} g(z)+\sum_{k=1}^K\sum_n \frac{c_{n,k}}{a_n^k}(\sum_{j=0}^J b_{k,j}(z/a_n)^{-j}+O((z/a_n)^{-J-1})$$
$$ =  z^{-m} g(z)+\sum_{j=-m}^J d_{J,j} z^{-j}+O(z^{-J-1}), \qquad uniformly \ as\ |z|\to \infty \ for \  arg(z)\in [-\pi+\delta,\pi-\delta]$$
This asymptotic expansion can be integrated and differentiated termwise to give the asymptotic of $f^{(m)}$.
