
> 
> Let $G$ be a finite subgroup of $\mathrm{GL}\_n(\mathbb C)$, $A,B \in G$ whose eigenvalues are thus in the unit circle.
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> 
> Assume that the eigenvalues ​​of $A$ are included in a circle arc of length $<\frac{\pi}3$, and the eigenvalues ​​of $B$ are included in a circle arc of length $<\pi$.
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> 
> Then $AB=BA$
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> 
> 


I have no idea how can I prove this, I had seen this result before on few books in which it was stated that it was a conjecture so I feel it's actually a very difficult problem for MSE.


If someone has any reference or proofs for this, it will be very helpful to share it.



Properties to have matrices that commute in $\mathrm{GL}_n(\mathbb C)$
This is a Theorem of Frobenius, whose proof can be found in the book of Isaacs on Character theory (14.15). 


There are several results of this type, including theorems of H. Blichfeldt. These are connected to proving Jordan's theorem about finite subgroups of ${\rm GL}(n,\mathbb{C}).$ There is an interesting section about these theorems in the 1962 book by Curtis and Reiner. See also Part A of L. Dornhoff's books on Representation Theory.


Most texts would treat the primitive (irreducible) linear group case. But the result reduces easily to the primitive case (we can assume the representation is irreducible (and unitary if desired)).
If the representation is imprimitive and induced (up to equivalence) from $H <G,$ then the conditions on $A,B$ imply that all $G$-conjugates of both $A$ and $B$ lie in $H.$ A Theorem of Blichfeldt and some use of Clifford's Theorem then force $A$ to commute with all its $G$-conjugates. Hence the $G$-conjugates of $A$ generate an Abelian normal subgroup of $G,$ say $K.$ Let $\lambda$ be an irreducible constituent of the representation of $K,$ and $I$ be its inertial subgroup. Then the representation of $G$ is induced from one of $I.$ As above, $B$ lies in every $G$-conjugate of $I.$ This means that $[A,B] \in {\rm ker}(\lambda^{g})$ for every $g \in G,$ so that $A$ and $B$ commute. So this is a theorem, not a conjecture. 


Results of this nature also appear in some of my own papers on linear groups, including one Journal of Algebra paper called (strangely enough) "On Linear Groups", and a later Journal of Algebra paper called "Further remarks on linear groups".


Updated edit: I have been doing further reading of my own papers on this subject. It is in fact the case that Theorem C' of my "On linear groups" paper shows that the result of the question has a positive answer even if the arc containing the eigenvalues of $A$ is allowed to have length less than $\frac{2 \pi}{5}$ (but the arc containing the eigenvalues of $B$ still has length less than $\pi$). It is also explicitly noted in my paper that Frobenius's theorem mentioned at the beginning answers the original question. 


The $\frac{2 \pi}{5}$ result can't be improved in general, as can be seen in ${\rm SL}(2,5).$
This last group has a two dimensional representation in which there is an element $A$ of order $5$ with spectrum $\{e^{\frac{4\pi i}{5}},e^{\frac{6 \pi i}{5} }\},$ such that $A$ does not commute with all its conjugates. 


It is sometimes possible to obtain sharper results if restrictions are placed on the prime divisors of the order of $A,$ and this is what is done some of the results of my "Further remarks on linear groups" paper (though note that there is a corrigendum published correcting some incorrect statements from "On linear groups" and "Further remarks on linear groups". This does not affect any statements made in this answer).


