Absolute value of polynomial

Question

I don't seem to grasp why

$$|1+6ωi-9ω^2| = 1+9ω^2, ω\gt 0$$

Where does the $6ω$ go? I'm thinking of $|6ωi|=6ω$.

Answer 1

Hint: Rewrite $-9\omega^2$ as $(3i\omega)^2$ and $6i\omega$ as $2(3i\omega)$. Can you take it from there?

Answer 2

It’s just the Pythagorean theorem applied to the real and imaginary parts of the number:

$$\begin{align*} |1+6\omega i-9\omega^2|&=\sqrt{(1-9\omega^2)^2+(6\omega)^2}\\ &=\sqrt{1-18\omega^2+81\omega^4+36\omega^2}\\ &=\sqrt{1+18\omega^2+81\omega^4}\;, \end{align*}$$

and from here you should have no trouble finishing.