Argument of complex number: what am I doing wrong?

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

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

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

What am I doing wrong?

• Take $\omega=\omega_0=1$, then $z=-i$. Which answer is correct?
– A.Γ.
Commented Jan 8, 2021 at 17:02

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 $$\arg z$$.
• Hi! Thank you for your answer. However I don't think I'm applying the property you mentioned, but rather $$\arg(z_1\times z_2)= \arg z_1+ \arg z_2$$ and $$\arg(z_1/ z_2)= \arg z_1- \arg z_2$$. Isn't this correct? Commented Jan 8, 2021 at 15:34