Finding the modulus of a complex number that satisfies a polynomial relation

Question

Consider $z\in\mathbb{C}$ such that $z^2-2z+3=0.$ Find the modulus of $$f(z)=z^{17}-z^{15}+6z^{14}+3z^2-5z+9$$

My attempt: $z^2-2z+3=0\Leftrightarrow\left[ \begin{array}{l}z=1+\sqrt{2}i\\z=1-\sqrt{2}i\end{array}\right.$

We have: $f(z)=\left(z^2\right)^7\times z^2-\left(z^2\right)^7\times z+6\left(z^2\right)^7+3z^2-5z+9\\=\left(z^2\right)^7\left[z^2-z+6\right]+3z^2-5z+9$

I do not know how to continue, please help me.

Answer 1

Hint: use polynomial division to write

$$f(z) = z^{17}-z^{15}+6z^{14}+3z^{2}-5z+9 = q(z)(z^{2}-2z+3) + r(z)$$

Since $z^{2}-2z+3 = 0$, then we have $f(z) = r(z)$. Since $r(z)$ should be linear in $z$, it should be easy to find the modulus from there. Feel free to comment if you need more help.

Answer 2

Here $z$ is such as $z^2-2z+3=0$ $$3(z^2-2z+3)=0 \;\; so\;\;3z^2-5z+9 = z$$ $$z^{17}=2z^{16}-3z^{15}=2(2z^{15}-3z^{14})-3z^{15}=z^{15}-6z^{14}$$ so$$f(z)=z$$