Real is hiding

As you might have heard your teachers saying "the argument of a complex number behaves like the log function".I'll use that same result.(But if you don't know why that happens I'd request you to get your doubt cleared)

$2arg(z^{1/3})=arg(z^2+\bar{z}z^{1/3})$ or, $arg(z^{2/3})=arg(z^2+\bar{z}z^{1/3})$ or, $(arg(z^2+\bar{z}z^{1/3})-(2arg(z^{1/3})$ =0 or, $(arg(z^2+\bar{z}z^{1/3})/(z^{2/3}))$ =0 or, $Im(arg[(z^2+\bar{z}z^{1/3})/(z^{2/3})])=0$

Now we use the result $Im(z)=(z-\bar{z})/2i$ and simplify.