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.

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.)
