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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 ?

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up vote 3 down vote accepted

$z = r e^{i\theta}$. Then you want to prove that $$\left \vert e^{i \theta} - 1\right \vert \leq \theta$$ $$e^{i \theta} - 1 = \cos(\theta) + i \sin(\theta) - 1 = -2\sin^2(\theta/2) + 2i \sin(\theta/2) \cos(\theta/2) = 2i \sin(\theta/2) e^{i \theta/2}$$

Hence, $$\vert e^{i \theta} - 1 \vert = 2 \left \vert \sin(\theta/2) \right \vert \leq 2 \times \left \vert \theta/2 \right \vert= \left \vert \theta \right \vert$$

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Thank you for your answer (and for leaving me some of the work!). but as I understand it $\theta$ is not unique in this representation. so I'm a bit confused...everything I write in the LHS can ba changed from $\theta$ to $\theta+2\pi k$ , is it OK ? – Belgi Oct 30 '12 at 22:06
@Belgi It actually doesn't matter since $\vert \sin(\phi) \vert \leq \vert \phi \vert$ for all $\phi$ – user17762 Oct 30 '12 at 22:09
Maybe the $\theta /2$ in the exponent in the last step in the big equation should have a minus sign ? – Belgi Oct 30 '12 at 22:18
@Belgi: I think it is fine, since $$-2\sin^2(\theta/2) + 2i \sin(\theta/2) \cos(\theta/2) = 2i^2\sin^2(\theta/2) + 2i \sin(\theta/2) \cos(\theta/2)$$ – user17762 Oct 30 '12 at 22:22
Thank you very much for your help! I do have to ask this - it seems that the proof is OK for every $\theta$ in $arg(z)$ but the LHS is fixed and the LHS can be as large as I there must be something that I don't undetstand... – Belgi Oct 30 '12 at 22:29

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