Let us consider \begin{equation} \lim_{x\to\infty}\left(\frac{3+i a-i/2}{3+i a+i/2}\right)^{2x} \end{equation} where $a\in\mathbb{R}$.
Depending on the sign of $a$, the limit is either 0 (for $a>0$) or infinity (for $a<0$) which one can obtain by writing the expression $\left(\frac{3+i a-i/2}{3+i a+i/2}\right)^{2x}=e^{2x\ln\left(\frac{3+ia-i/2}{3+ia+i/2} \right)}$ and looking at the sign of the real part of the exponent.
Now, I would like to look at the limit \begin{equation} \lim_{x\to\infty}\left(\frac{\sinh(3+i a-i/2)}{\sinh(3+i a+i/2)}\right)^{2x}=\lim_{x\to\infty}e^{2x\ln\left(\frac{\sinh(3+i a-i/2)}{\sinh(3+i a+i/2)}\right)} \end{equation} where $a\in\mathbb{R}$
Here, looks like the sign of the exponent does not depend just on the sign of $a$.
Since, $\sinh$ is a monotonic function, I was expecting the result to be the same as previous case.
How do I go about finding all the values of $a$ for which the limit vanishes and all the values of $a$ for which the limit is infinite?
Use the identity $ \sinh(z) = \frac{e^z - e^{-z}}{2} $ to get \begin{align} f(a)&:=\frac{\sinh(3+i a-i/2)}{\sinh(3+i a+i/2)}\\ &= \frac{e^{3+i a-i/2} - e^{-3-i a+i/2}} {e^{3+i a+i/2} - e^{-3-i a-i/2}}\\ &= \frac{e^{6+2i a} - e^{i}} {e^{6+2i a+i} - 1} \\ &= \frac{e^{6+2i a} - e^{i}} {e^{6+2i a+i} - 1} \cdot \frac{e^{6-2i a-i} - 1}{e^{6-2i a-i} - 1} \\ &= \frac{e^{12-i} - e^{6+2ia} - e^{6i+2a+1} + e^i} {e^{12}+1 - e^{6+2i a+i} - e^{6-2i a-i}} \\ &= \frac{e^{12}[cos(1)-i\sin(1)] - e^{6+2ia} - e^{6i+2a+1} + e^i} {e^{12}+1 - 2e^{6} \cos(2a+1)}.\tag{*}\label{*} \end{align} Since $e^{12}$ is dominates the other terms, the denominator in the obtained expression is positive, and the nominator has positive real part $e^{12}\cos(1)>0.5 e^{12} > e^6$.
Next, we use the fact the basic fact about complex numbers that for $u \in \mathbb C$ $$ \lim_{x\to \infty} u^{2x} $$ converges if $|u|<1$ or if $u=\pm 1$, and diverges otherwise. It only remains to estimate the absolute value of the nominator and denominator of the formula \eqref{*}. We have $$ \big|e^{12}[cos(1)-i\sin(1)] - e^{6+2ia} - e^{6i+2a+1} + e^i \big| < e^{12} \cos(1) + 3 e^6, $$ and $$ \big| e^{12}+1 - 2e^{6} \cos(2a+1) \big| > e^{12} - 3 e^6. $$ However, as $\cos(1) \approx 0.54$ and $e^{12}/e^{6} = e^6 \approx 403$, $$ e^{12} \cos(1) + 3 e^6 < e^{12} - 3 e^6. $$ We conclude that $$ |f(a)| = \left| \frac{\sinh(3+i a-i/2)}{\sinh(3+i a+i/2)} \right| < 1, $$ and so the sequence $(f(a))^{2x}$ converges to 0 as $x\to \infty$ for any $a\in \mathbb R$.
Back to the first problem: We want to know if $(g(a))^{2x}$ converges as $x\to\infty$, where $$ g(a) = \frac{3+i a-i/2}{3+i a+i/2}. $$ We have $$ |3+i a \pm i/2|^2 = 9 + (a\pm 1/2)^2 = 9 + a^2 + \tfrac 1 4 \pm a, $$ and so $|g(a)|<1$ if and only if $a>0$. (If $a=0$ $|g(a)|=1$, but it is not equal $\pm 1$.) We conclude that $(g(a))^{2x}$ converges (to $0$) if and only if $a>0$.