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Could anyone help me find this limit $$\limsup_{|z|\to\infty}\frac{\log|e^{-iz}|}{|z|}$$

where $z=x+iy, x,y\in \mathbb R$.

I guess we need to use that $e^{-iz}=\cos z- i\sin z$, then $$\limsup_{|z|\to\infty}\frac{\log|e^{-iz}|}{|z|}=\limsup_{|z|\to\infty}\frac{\log|\cos z- i\sin z|}{|z|}$$

but what next?

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    Better use $|e^z| = e^{\text{Re}(z)}$.2012-04-18
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    @Dirk I nicer template of the `Re` symbol is obtained by $\Re$ (`\Re`). I guess you might also use $\mathfrak{Re}$ (`\mathfrak{Re}`)2012-04-18
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    Seems to be a matter of taste and habit. I feel like $\text{Re}$ is more popular that $\Re$ in Germany...2012-04-19

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First, because $e^{ib}$ has modulus (absolute value) $1$ for any real $b$, and $e^x$ is always positive for any real $x$, we have the following identity for any $a\in\Bbb C$:

$$\log |e^a|=\log |e^{\mathrm{Re}(a)}e^{i\mathrm{Im}(a)}|= \log e^{\mathrm{Re}(a)}=\mathrm{Re}(a).$$

Thus, writing $z=x+iy$, you need to evaluate

$$\limsup_{|z|\to\infty} \frac{\mathrm{Re}(-iz)}{|z|}=\limsup_{|z|\to\infty} \frac{y}{\sqrt{x^2+y^2}}=\limsup_{|z|\to\infty} \, (\sin\theta).$$

Where $\theta$ is the argument of $z$. Note for any circle $|z|=R>0$ in the complex plane, there is a $z$ for any given angle $\theta\in[0,2\pi)$. Consider the positive imaginary axis...

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    So I think the last limit does not exit!2012-04-18
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    @Megan: Why do you think that?2012-04-18
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    Sorry, it is exist and equals to 1. Correct me if I'm wrong!2012-04-18
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    @Megan: That's correct! The normal limit certainly doesn't exist, but the limsup does.2012-04-18
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    Taking the limit along the imaginary axis gives us 1, but is that enough? How I can prove it in general, in any direction?2012-04-18
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    @Megan: What do you mean by "any direction"? You aren't supposed to fix the direction; are you familiar with limsup? No matter how big $R$ is, $\theta=\pi/2$ attains the global maximum of $1$. It does not have to be the case that other directions also attain this, nor is it the case.2012-04-19
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    One could rewrite it as $$\limsup_{|z|\to\infty} \sin\theta = \lim_{R\to\infty}\sup_{|z|>R}\sin\theta $$ for easier evaluation.2012-04-19
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    I like that the expression has been replaced by sin, and I think that the result is most clear this way. However, an alternative approach would be to consider the radial limits in every direction directly by setting y=mx and letting m range over the reals. For all finite m, the limit is less than one. By considering the imaginary axis separately, we see that the expression attains its maximum value of 1 at every purely imaginary number.2012-04-19