$\text{arg}(i \: \text{conj}(z))=\text{arg}(i)+\text{arg}(\text{conj}(z))$ This is a Complex Analysis question.
Let z be a complex number, i be the imaginary unit, arg be the argument of z, and $\bar{z}$be the complex conjugate of z.
How do I prove that the following equality holds?
$\text{arg}(i \: \bar{z})=\text{arg}(i)+\text{arg}(\bar{z})$
I haven't been able to prove it using the definition of the argument function together with the trig form of a complex number alone. Please do not prove it using the polar form if possible.

For $z=a+bi$, use $\arg(z_1z_2)=\arg(z_1)+\arg(z_2),\,\mod(-\pi,\pi]$:
$$
\arg(i\overline{z})=\arg(i)+\arg(\overline{z}).
$$

To see why the identity is true, note that $z=|z|e^{i\arg z}$, so
$$
\arg(z_1z_2)=\arg(|z_1z_2|e^{i(\arg(z_1)+\arg(z_2))})=\arg(z_1)+\arg(z_2),\,\mod (-\pi,\pi].$$
Throughout I am using $\arg$ as the principal argument, since it seems LaTeX doesn't have $\Arg$ built in (?).
