Analytical Function at Real infinity

Question

Let's consider a function $f(z)$ analytical everywhere except at infinity. It's zero at real infinity $\lim_{x\rightarrow \infty}f(x)=0$.

Is it true that $\lim_{x\rightarrow \infty}f(x+iy)=0$, where $x,y$ are reals?

Answer 1

$\lim_{x\to \infty}f(x)=0$ does not imply $\lim_{x\to \infty}f(x+iy)=0$ for $y \in \Bbb R$, a counterexample is $$ f(z) = \frac{\sin(z^2)}{z} \, . $$ for $z \ne 0$, $f(0) = 0$. $f$ is an entire function. Using $$ \sin(x+iy) = \sin (x) \cosh (y) + i \cos (x) \sinh (y) $$ we get $$ | \sin(x+iy) |^2 = \sin^2(x) + \sinh^2(y) $$ and then $$ |f(x+iy)|^2 = \left | \frac{\sin((x^2 - y^2) + i (2xy))}{x+iy} \right |^2 = \frac{\sin^2(x^2-y^2) + \sinh^2(2xy)}{x^2+y^2} $$ For $y=0$ we have $$ |f(x)|^2 = \frac{\sin^2(x^2)}{x^2} \le \frac{1}{x^2} \to 0 $$ for $x \to \pm \infty$. But for any $y \ne 0$ $$ |f(x+iy)|^2 \ge \frac{\sinh^2(2xy)}{x^2+y^2} \to \infty $$ for $x \to \pm \infty$ because $\sinh$ grows exponentially, and therefore faster than any polynomial.