System of equations in a,b,c,d $a,b,c,d$ are complex numbers satisfying
\begin{cases}
  a+b+c+d=3 \\
  a^2+ b^2+ c^2+ d^2=5 \\
  a^3+ b^3+ c^3+ d^3=3 \\
  a^4+ b^4+ c^4+ d^4=9
\end{cases}
Find the value of the following: $$a^{2015} + b^{2015} + c^{2015} + d^{2015}$$.

$a^2+b^2=(a+b)^2-2ab=(3-(c+d))^2-2ab \implies ab=2-3(c+d)+(c+d)^2-cd$
$a^3+b^3=(a+b)^3-3ab(a+b)=(3-(c+d))^3-3(3-(c+d))(2-3(c+d)+(c+d)^2-cd)$
$c^3+d^3=(c+d)^3-3cd(c+d)$
$u=c+d,v=cd \implies (3-u)^3-3(3-u)(2-3u+u^2-v)+u^3-3uv=3 \implies v=\dfrac{u^3-3u^2+2u+2}{2u-3} $
$a^4+b^4+c^4+d^4=(a^2+b^2)^2-2(ab)^2+((c+d)^2-2cd)^2-2(cd)^2=(5-u^2+2v)^2-2(u^2-3u+2-v)^2+(u^2-2v)^2-2v^2=9$
$v^2-u^2v-3uv+7v+3u^3-9u^2+6u+2=0$ ,replace $v$ we get:
$(u-2)(u-1)(u^2+1)(u^2-6u+10)=0$
$u_1=1,u_2=2,u_3=i,u_4=3-i,u_5=-i,u_6=3+i$
note: $u_1+u_2=u_3+u_4=u_5+u_6=3$ which means the root is simply different combination of $a,b,c,d$, so we only take one root is enough for final answer,take $u=1 \implies v=-2\implies c=-1,d=2 ,a=1+i,b=1-i$
so the final answer is easy to calculate.
