Consider a group $G$ with exactly 4 conjugacy classes $C_1,\;C_2,\;C_3\;C_4$ containing exactly 1, 3, 4, 4 elements, respectively. Let $\zeta$ be a primitive third root of 1 and the following table its character table. Complete it.
My attempt: The group has 4 irreducible representations. Let $\chi_0, \chi_1, \chi_2, \chi_3$ the corresponding irreducible characters. Using the character of the regular representation I found that $\chi_0(1)=\chi_1(1)=\chi_2(1)=1$ and $\chi_3(1)=3$.
Now, since each element of $C_2$ has order 2, then $\chi_0(g)=\chi_1(g)=\chi_2(g)=\chi_3(g)=1$, where $g$ is an element of $C_2$.
Also using that $1=<\chi_2,\chi_1>$, I found that $\chi_2(g)=2-\zeta$, where g is an element of $C_4$.
Therefore, so far the table goes like this.
Problem: I don't know how to find the other four values. I have tried with the orthogonality rules but I have not been able to eliminate any of the 4 unknowns.