For the hyperbolic tan function, we have the property
$\displaystyle \lim_{a \to \infty} \tanh(a(z-b)) = \mathrm{signum}(z-b)$
where $z$, $a$ and $b$ are real.
But what happens if $b$ is complex (say, $b=b_1+ib_2$)? Does the above relation still hold and how can we prove it?
Using the angle-addition formula $$\tanh{(A+B)} = \frac{\tanh{A}+\tanh{B}}{1+\tanh{A}\tanh{B}} = \frac{\sinh{A}\cosh{B}+\sinh{B}\cosh{A}}{\cosh{A}\cosh{B}+\sinh{A}\sinh{B}}, $$ we can expand the hyperbolic tangent. The real part of $b$ makes no difference, because we can just set $x=z-b_1$ to remove it, so suppose $b=ic$. Then $$ \tanh{a(x-ic)} = \frac{\sinh{ax}\cosh{(-iac)}+\sinh{(-iac)}\cosh{ax}}{\cosh{ax}\cosh{(-iac)}+\sinh{ax}\sinh{(-iac)}}, $$ and $\sinh{(-iz)}=i\sin{z}$, $\cosh{(-iz)}=\cos{z}$, so we have $$ \tanh{a(x-ic)} = \frac{\sinh{ax}\cos{ac}+i\sin{ac}\cosh{ax}}{\cosh{ax}\cos{ac}+i\sinh{ax}\sin{ac}}, $$ Multiplying this by $1$ in the form $(\cosh{ax}\cos{ac}-i\sinh{ax}\sin{ac})/(\cosh{ax}\cos{ac}-i\sinh{ax}\sin{ac})$, and then using the Pythagorean and double-angle formulae, we find eventually that $$ \tanh{a(x-ic)} = \frac{\sinh{2az}-i\sin{2ac}}{\cosh{2az}+\cos{2ac}}. $$ Here, we know that the trigonometric terms are bounded by $1$, so the limiting behaviour is dominated by the limit of $\sinh{2az}/\cosh{2az}=\tanh{2az}$, which tends to $\sgn{(z)}$ as $a\to +\infty$. So the answer is yes, the formula is valid for complex $b$.