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?

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

1 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 :)

  • $\begingroup$ 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? $\endgroup$ Commented Jan 8, 2021 at 15:34
  • $\begingroup$ @GrangerObliviate: Can you then elaborate how you found the argument for numerator? $\endgroup$ Commented Jan 9, 2021 at 17:38

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