Finding the argument of a complex function I've the following transfer function:
$$H(s)=\frac{1}{as^3+bs^2+cs+1}$$
Where $a,b,c$ are all real and positive.

How can I find $\arg(H(i\omega))$? And I know that $\omega\ge0$


What I did:
$$H(i\omega)=\frac{1}{a(i\omega)^3+b(i\omega)^2+c(i\omega)+1}=\frac{1}{-a\omega^3i-b\omega^2+c\omega i+1}=$$
$$\frac{1}{1-b\omega^2+(c\omega-a\omega^3)i}$$
Now finding the argument I can write:
$$\arg(H(i\omega))=\arg(1)-\arg(1-b\omega^2+(c\omega-a\omega^3)i)=$$
$$0-\arg(1-b\omega^2+(c\omega-a\omega^3)i)=-\arg(1-b\omega^2+(c\omega-a\omega^3)i)$$
Now, how can I setup a function that depends on the value of $a,b,c,\omega$?

Denote $\theta=\arg(H(i\omega))$ and $$r=|1-b\omega^2+(c\omega-a\omega^3)i|=\sqrt{(1-b\omega^2)^2+(c\omega-a\omega^3)^2}.$$
From $$\theta=-\arg\big(1-b\omega^2+(c\omega-a\omega^3)i\big)$$ we get $$r\cos \theta =1-b\omega^2 \quad \text{and} \quad r\sin \theta =-(c\omega-a\omega^3)$$
and then $$\tan \theta = \frac{a\omega^3-c\omega}{1-b\omega^2}.$$ Finally $$\theta = \arctan\frac{a\omega^3-c\omega}{1-b\omega^2}.$$
I assumed silently that $1-b\omega^2\neq 0.$ If it were $0,$ the argument would be clear since the beginning.
