I am trying to solve the following:
Let $s$ be a real number. For what $s$ is $\dfrac{(-1)^s-1}{i^s}$ a Gaussian integer?
But I feel this question sits far beyond my ability. I do not know how to start; any help or hint where to begin would be appreciated.
On
Assuming that $i^s$, with $s\in \mathbf R$, has been correctly defined, it is straightforward that $\frac {(-1)^s - 1}{i^s}=\frac {i^{2s}- 1}{i^s}=i^s-i^{-s}$, and we are brought back to the determination of the real numbers $s$ for which $i^s-i^{-s}\in \mathbf Z[i]$. Allow me to recall some classical facts from complex analysis (cf. any text book, e.g. H. Cartan's "Elementary theory of analytic functions...", I-5 and II-7). For $z\in \mathbf C$, the exponential function exp$(z)$ is defined by the usual power series, which converges absolutely everywhere in $\mathbf C$. The problem is with the "reciprocal" logarithm function log$(z)$, which is "multiform" because exp$(2k\pi i)=1$ for $k\in \mathbf Z$. So one begins to define a determination of log$(z)$ on an open connected domain $D\subset \mathbf C$, not containing $0$, as a continuous function f$(z)$ s.t. exp(f$(z))=z$ for all $z\in D$. Clearly, because $D$ is connected , two determinations of log$(z)$ on $D$ differ by a multiple of $2i\pi$. The existence of such determinations is not a trivial affair. One must appeal to homotopy to show that, if $D$ is simply connected, then a determination of log$(z)$ can be defined on $D$. In this case, for any $t\in\mathbf C$ and any $z\in D$, one can also define $z^t$:= exp($t$log($z$)).
Back to your problem, on an open simply connected domain $D$ not containing $0$, one can define $i^s$= exp($s$log($i))$, with $2$ log$(i)$ = log$(-1)$ = log (exp$(i\pi))=i\pi$, so that $i^s-i^{-s}=2i$sin$(s\pi/2)$, which belongs to $\mathbf Z[i]$ iff sin$(s\pi/2) \in {\frac 12}\mathbf Z$, iff sin$(s\pi/2)=0,\pm \frac 12$ .
Hint: Let $$(-1)^s-1=i^s\cdot (a+bi).$$ Then triangle inequality gives a serious constraint.