What function satisfies f(0)=1,f(1)=−i2,f(2)=−1,f(3)=−if(0)=1,f(1)=−i^2,f(2)=−1,f(3)=−i?

    Assuming it is a cubic function i=(i)^(0.5) f(0)=1=-(i^2)^0 f(1)=-i^2=-(-1)=1=-(i^2)^1 f(2)=-1=-(i^2)^2 f(3)=-1(i^2)^3 b=1 let f(x)=a(x)^3+b(x)^2+c(x)+d f(0)=1=d f(1)=1=a+b+c+1=>a+b+c=0 f(2)=-1=8a+4b+2c+1 6a+2b=-2 3a+b=-1=> 24a+8b=-8 f(3)=27a+9b+3c+1=-i 24a+6b=-i-1 2b=i+7 b=(i+7)/2 a=-(1+b)/3=-(i+9)/6 c=-(a+b)=(-i-9+3i+21)/6=(2i+12)/16=(i+6)/8 d=1 f(x)=-((i+9)/6)x^3+((i+7)/2)x^2+((i+6)/8)x+1