If $a$ and $b$ are the roots of $z^2 - 2z + 4 = 0$ then what is $a^n + b^n + ab$ ($n$ is a natural number)?

Question

I don't know how to solve this question, any help would be appreciate it.

If $z^2 - 2z + 4 = 0$, then what is the result of this $a^n + b^n + ab$ ($n$ is a natural number, $a$ and $b$ are the roots of that equation)?

Answer 1

HINT:

$$z=1\pm\sqrt3i=2\left(\cos\dfrac\pi3\pm i\sin\dfrac\pi3\right)$$

Using de Moivre's formula,

$$z^n=2^n\left(\cos\dfrac{n\pi}3\pm i\sin\dfrac{n\pi}3\right)$$

Answer 2

Hint 1: You know that $a^2=2a-4$. Thus $a^3 = 2a^2-4a = \dots$

Hint 2: You know that $z^2-2z+4 = (z-a)(z-b)=z^2-(a+b)z+ab$.

Answer 3

HINT:

$$z=1\pm\sqrt{3}i=-2w,-2w^2$$ where $w$ is a complex cube root of $1$

$\implies ab=4$

Need to deal $\displaystyle n\equiv0,1,2\pmod3$ separately.