$$\arg\left(\frac{1}{z}\right) = \arg(\bar z)$$
So, I used the definition $z\bar z = |z|^2$
Then I divided both sides by $z$; $$\bar z= \frac{|z|^2}{z}$$But $|z|^2$ is a scalar, $>0$
Then $$\arg\left(\frac{|z|^2}{z}\right) = \arg(\bar z)$$
This is where I am stuck. Is $|z|^2$ a constant? Do I take it out and write it as$|z|^2\arg\left(\frac{1}{z}\right)$ or no; what do I do?
$\endgroup$ 23 Answers
$\begingroup$HINT
Polar form of a complex number $$z=re^{i\theta}$$
$\arg(z)=\arg(re^{i\theta})=\theta$
We don't take the constant out but we ignore it for we are only interested in the angle when we write $\arg$.
$\endgroup$ $\begingroup$Suppose that $z \neq 0$ is given in the polar form as $z = r e^{i\theta}$.
We have
$$\arg(1/z)= \arg(\frac{e^{-i\theta}}{r} )= -\theta = \arg(re^{-i\theta}) = \arg(\overline z )$$
$\endgroup$ $\begingroup$Following your approach, whenever $c > 0 $ , then $\arg(cz)=\arg(z) $
Since $ |z|^{2} $ a real constant $\gt $ 0 ,so, $\arg(\frac {|z|^{2}}{z} ) = \arg(\frac{1}{z}) $.
So, $\arg(\bar z ) = \arg(\frac{1}{z}) $ ( as, $\bar z = \frac{|z|^{2}}{z} $)
