Argument of complex number: what am I doing wrong?

Question

Find $\arg z$ where $$z=-\frac{i\omega-w_0}{i\omega+w_0}$$ where $\omega$ and $\omega_0$ are positive numbers.

My attempt was this:

$$\arg z=\pi +\arctan\left(\frac{\omega}{-\omega_0}\right)+\pi-\arctan\left(\frac{\omega}{\omega_0}\right)$$$$=2\pi -\arctan\left(\frac{\omega}{\omega_0}\right)-\arctan\left(\frac{\omega}{\omega_0}\right)$$$$=2\pi-2\arctan\left(\frac{\omega}{\omega_0}\right)$$

However, the correct answer seems to be: $$\arg z =\pi-2\arctan\left(\frac{\omega}{\omega_0}\right)$$

What am I doing wrong?

Answer 1

You seem to have misunderstood concepts. $$\arg(z_1\pm z_2)\neq \arg z_1\pm \arg z_2$$ The proper and intended way would be to write $z$ in the form $a+bi$. This can be done as follows: $$z=\frac{\omega_0-i\omega}{\omega_0+i\omega}=\frac{(\omega_0-i\omega)(\omega_0-i\omega)}{(\omega_0+i\omega)(\omega_0-i\omega)}$$ $$z=\frac{\omega_0^2-\omega^2-2\omega_0\omega i}{\omega_0^2+\omega^2}$$ $$z=\frac{\omega_0^2-\omega^2}{\omega_0^2+\omega^2}+\frac{-2\omega_0\omega}{\omega_0^2+\omega^2}i$$ And with this, we can easily calculate $\arg z$.

Hope this helps. Ask anything if not clear :)