I need some help with the following problem:
Let $0\neq z\in\mathbb{C}$, prove $\lvert\frac{z}{\lvert z\rvert}-1\lvert\leq\rvert Arg(z)\rvert$
where we take the argument to be in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right]$.
What I did:
I say that it is clear that it sufficient to prove that if $z'$ is on the unit circle then $|z'-1|\leq|Arg(\alpha z)|$ where $0\neq\alpha\in\mathbb{R}$.
Since if $0\neq\alpha\in\mathbb{R}$ then $Arg(\alpha z)=Arg(z)$ we need to prove that $|z'-1|\leq|Arg(z)|$.
Now, let $a,b\in\mathbb{R}$s.t $z'=a+bi$ then $\sqrt{a^{2}+b^{2}}=1$ and we need to prove that $\sqrt{(a-1)^{2}+b^{2}}\leq|Arg(z')|$.
This is where I am stuck, can someone please provide a hint ? am I even on the right track ?