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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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    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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    I like that the expression has been replaced by sin, and I think that the result is most clear this way.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