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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 ?

1 Answers 1

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$z = r e^{i\theta}$. 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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    That is, write $z=a+bi$ then the LHS is some fixed number expressed with $a,b$. the LHS is unbounded if we take some large value of $k$ in $\theta +2\pi k$...2012-10-30