$f(z)$ is a complex polynomial, and its remainders with $z^2+z+1$ and $z-1$ are given to be $2wz+w$ and $5-w$ respectively. Given that $z^3-1=(z-1)(z^2+z+1)$ and $w = \frac{-1+i\sqrt{3}}{2}$, find the remainder when $f(z)$ is divided with $z^3-1$.
My approach to this:
- $f(z) = (z^2+z+1) \times q_1(z)+(2w z+w)$
- $f(z) = (z-1) \times q_2(z) + (5-w)$
Replacing $z$ with $w$ and $w^2$ in equation 1 (as roots of $z^2+z+1$ are $w$ and $w^2$), we get:
- $f(w) = 2wz+w$ and
- $f(w^2)=2wz+w$
(Basically, applying remainder theorem from teh basics)
Similarly, replacing $z$ with 1 in equation 2, we get:
- $f(1)=5-w$
However, I'm not able to go any further. Is there any mistake in my approach? And can someone please give me a hint to go forward?
Note that division by $z^3 -1$ gives as remainder a polynomial of degree at most 2, call it $r(z) = az^2 + bz + c$ where $a, b, c \in \mathbb{C}$. Write $$f(z) = (z^3 - 1)q(z) + r(z)$$ Because $1, w, w^2$ are roots of the polynomial $(z^3-1)$ (i.e. substituting any of these numbers into the polynomial gives 0), we have $$ f(1) = 5-w =(1^3-1)\cdot q(1) + r(1) = r(1) = a + b + c$$ $$f(w) = 2w^2 + w = r(w) = aw^2 + bw + c $$ $$f(w^2) = 2 + w^2 = r(w^2) = aw^4 + bw^2 + c$$ Can you finish by solving for $a, b, c$?