I came across the following statement while reading the proof of a fact involving the branch of logarithm:
Since $|z|<1$ we get that $\arg(1 \pm z)$ lies between $-\dfrac{\pi}{2}$ and $\dfrac{\pi}{2}$.
I know that the (principal) argument $\arg(z)$ belongs to $(-\pi,\pi]$. However, unable to see how $|z|$ can affect the principal argument of $1-z$.
Also, when checking WolframAlpha for some properties of $\arg(1-z)$, I came across this:
Now I feel even more confused. Kindly let me know if you have any hints for solving this problem. I feel like I am missing something pretty obvious.
$|z|=x^2+y^2<1$ describes an open disc of radius $1$ centered at $0$. Clearly, $-1<x<1$, however $|z+1|$ is an open disc translated in the real axis by $1$, in other words, $0<x<2$. Since $x$ is positive, then the argument of $1+z$ lies between $-\pi/2$ and $\pi/2$. For the other case add $\pi$, which will yield the same result