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In Roger Penrose's book Road to Reality - Chapter 5 - he goes to great lengths to arrive at the standard polar expression for a complex number $w = r e^{i \theta}$ via a discussion of complex logarithms, see page 94-95 of the following link.

The key step, on page 94, is to look for an inverse to $w = e^z$ to define $\log(w) = z$. Penrose denotes the radius and argument of $w$ as $[r,\theta]$ and then just states that the logarithm of $w$ - i.e. $z$ - is:

$\log(w) = z = \log(r) + i \theta$.

Penrose then uses the above equation $\log(z) = \log(r) + i \theta$ to 'arrive at the result' that $z = \exp(w) = \exp(\log r + i \theta) = r \exp(i\theta).$

My question is: is it internally consistent / possible to arrive at the standard polar form of a complex number from the logarithm as Penrose does, or is this just circular? Put another way, does it make sense to claim the form of $\log(w) = z =\log(r) + i \theta$ above without reference to $z = r \exp(i\theta)$?

Couple more notes / comments

  • Penrose points out in the text below the expression for $\log(w) = z=\log(r) + i \theta$ that it is " remarkable [] that the imaginary part of $z$ is just the angle $\theta$ that is the argument of the complex number $w$". There is a footnote to this that points out that he doesn't claim to have proven that the $i \theta$ term could not be a real multiple of $i \theta$ since this "requires calculus" - but I don't know what this means.

  • My approach to getting to the logarithm expression would be to to observe that $e^{i\theta} = \cos\theta + i\sin\theta$ via power series expansions and then note that this is directly the definition of $w$ by basic trigonometry (real part of $w$ equals $r \cos\theta$ etc. etc). The expression for $\log(w)$ then follows automatically by properties of logarithms. But Penrose has very clearly gone via the logarithm argument to get at the polar form of a generic complex number, and he's a much smarter man than me, so I can't help but feel I'm missing something profound in his argument. What am I missing?

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I can kind of try. I found this out once in my maths class, but my teacher put it off and didn't pay any attention to it.

$$Z=R(\cos\theta+i\sin\theta)\\ \frac{dZ}{d\theta}=R(-\sin\theta+i\cos\theta)=R×\frac{iZ}{R}=iZ\\\implies \frac{1}{Z}dZ=i\ d\theta$$ Taking integral on both sides, $$\ln Z=i\theta+C$$ Taking $\theta=0$, we can prove $C=\ln R$. Thus $$\ln Z=i\theta+\ln R$$ From here, we can prove that $$Z=Re^{i\theta}$$ I haven't learnt differentiation in complex plane yet. Anyone who thinks I've made a mistake please inform me so that I can rectify it :)

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  • $\begingroup$ Thanks for the input. I initially really liked this approach, but the more I think about it the more I think this may have snuck log(z) = log(|z|) + i Arg (z) in by the back door. Why? Because to state the anti-derivative of dZ/Z is log(Z) you need to know the derivative of log(Z) is 1/Z where Z is complex. And this requires showing it is differentiable at the specific point Z (or analytic in a region) which requires a specification of log(Z). To do this requires a slightly subtle argument because log(Z) is either multivalued or discontinuous - meaning you need a branch cut. $\endgroup$
    – a_former_scientist
    Commented May 17 at 8:15

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