In school, we learnt how to prove some of the basic properties of $\text{Arg}$, one of them being $$ \text{Arg}(z_1)+\text{Arg}(z_2)=\text{Arg}(z_1z_2) $$ We did this by writing $z_1$ and $z_2$ in the modulus-argument form, and using $e^{i\theta}=\cos\theta+i\sin\theta$. I noticed that $\text{Arg}(z)$ has a similar property to logarithms, in that $$ \log a+\log b = \log ab $$ Is this just a coincidence, or is there a deeper reason for this?
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2$\begingroup$ $r_1e^{ia}r_2e^{ib}=r_1r_2e^{i(a+b)}$ $\endgroup$– J. W. TannerSep 21, 2020 at 21:26
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$\begingroup$ If this is what your looking for, they are roughly the same $Arg (z)=\theta = \frac{\ln (e^{i \theta})}{i}$ $\endgroup$– Fareed Abi FarrajSep 21, 2020 at 21:30
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1$\begingroup$ Its not a coincidence. Try to write down 1) what is the definition of a logarithm and 2) what is the definition of the argument. This will help you see the connection. $\endgroup$– WintherSep 21, 2020 at 21:31
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2 Answers
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The imaginary part of $\log z$ is $\text{arg}$ z.
$$\log z = \log |z| + i \,\text{arg } z$$
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$\begingroup$ $\ln(z)=\ln(|z|e^{i\text{Arg}(z)})=\ln |z|+i\, \text{Arg}(z)$ $\endgroup$– AmadSep 21, 2020 at 22:04
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1$\begingroup$ We could use the notation $\ln z$ or $\log z$ both mean the same thing. Various textbooks use one or the other. The notation $\text{Arg }z$ is meant to specify a principal argument, with respect to a principal branch cut. The notation $\arg z$ works just as well, in this context, but does not specify which branch. By-the-way, the "principal" branch is also arbitrary. $\endgroup$– mjwSep 22, 2020 at 1:47
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$\begingroup$ @mjw hmm when I was studying, it was the opposite: $\operatorname{Arg}$ meant all branches, while $\arg$ only the primary branch. $\endgroup$– AnixxJun 23, 2021 at 15:05
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Each complex number can be written as $$z=|z|e^{iArg(z)}$$ $$e^{iArg(z)}=\frac{z}{|z|}$$ $$iArg(z)=\ln(\frac{z}{|z|})$$ $$Arg(z)=-i\ln(\frac{z}{|z|})$$ Looking at this you can see that argument of complex number is closely related to the logarythmic function.
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1$\begingroup$ You have to say that passing from the 2nd to the third line is only to convey the idea, because the logarithm (with an "i", pun intended) of a complex number has to be defined, and we know it deserves new definitions (cuts, etc....) $\endgroup$ Sep 21, 2020 at 21:37
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$\begingroup$ Yes, loosely speaking the $\ln$ and $Arg$ functions are both commonly defined with a branch cut at negative real numbers ($\theta=\pi$) so that the range of $iArg$ is from $-i\pi$ to $i\pi$ and together with the fact that $\frac{z}{|z|}$ lies on the unit circle, $\ln(\frac{z}{|z|})$ ranges from $-i\pi$ to $i\pi$ too $\endgroup$– AmadSep 21, 2020 at 21:56
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1$\begingroup$ This is a nice way of looking at it, starting with the exponential function. $\endgroup$– mjwSep 22, 2020 at 1:49