I believe I know how to do it, but first I have to know what you mean by $\sqrt{1-4ik}$. Usually $\sqrt x$ is defined for $x\geq 0$ and it is equal to the number $y\geq 0$ such that $x=y^2$. If you want to extend the definition to complex numbers you have to specify which of the two branches of the radical you consider. If I am to guess, then perhaps you choose $\sqrt{1-4ik}$ to be the complex number $z$ with $1-4ik=z^2$ and $\Re z\geq 0$. Is that right?

@ViditNanda Thank you for your answer, but I don't know how to use what is on page 283. I guess it has something to do with what Denis-Charles Cisinski wrote above, because in that section one can find the octohedral axiom. Unfortunately, I'm a noob in cohomology so I don't understand that. I only know some basics, like the standard resolution, long exact sequences, cup products etc. I never needed more than this. This particular result can be proved by rudimentary means in half a page, maybe a little more. I was looking for a quotable rudimentary proof.

Initially, for a morphism of $G$-modules $f:M\to M$ I defined $H^n(G,M,f)=Z^n(G,M,f)/B^n(G,M,f)$, where $Z^n(G,M,f)=\{ (a,b)\in C^n(G,M)\times C^{n+1}(G,M)\mid\, da=f(b),\, db=0\}$ and $B^n(G,M,f)$ is generated by the obvious elements of $Z^n(G,M,f)$ namely, $(dc,0)$, with $c\in C^{n-1}(G,M)$ and $(f(c),c)$, with $c\in C^n(G,M)$. This definition worked fine, until I realized that $H^n(G,M,f)$ is the cohomology of a complex. Now I cannot pretend I don't know about mapping cones and go by the old definition, but I wish I could keep the homological theory to a minimum.

@Denis-CharlesCisinski Thank you for your answer, but I would prefer a more low-tech approach. My paper is aimed at people from number theory, who are not usually experts in cohomology. In fact, I only learned about mapping cones two days ago, when I asked on mathoverflow. mathoverflow.net/questions/396916

@Wojowu Thank you. This was a very quick and helpful answer.