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In the Algebra chapter of the Feynman Lectures on Physics, Feynman introduces complex powers:

Thus $$10^{(r+is)}=10^r10^{is}\tag{22.5}$$ But $10^r$ we already know how to compute, and we can always multiply anything by anything else; therefore the problem is to compute only $10^{is}$. Let us call it some complex number, $x+iy$. Problem: given $s$, find $x$, find $y$. Now if $$10^{is}=x+iy$$ then the complex conjugate of this equation must also be true, so that $$10^{−is}=x−iy$$

I don't think it's all that easy to guess/infer intuitively, this fact (about complex conjugates). Especially when you are a beginner (that's who the author addresses this chapter to, building steadily from arithmetic through algebra, logarithms, etc... guided by the intellectual beacons of Abstraction and Generalisation), it isn't at all convincing why if $10^{is}$ equals some $x + iy$, then $10^{-is}$ must be $x-iy$.

The only definition of $i$ is that its square is $-1$ (Feynman's reason for this to be true). What am I missing here?

Ad N Khan
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  • Look in to complex analysis, Euler's formula, and logarithms. – Sean Lake Aug 03 '21 at 19:55
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    Who is claiming that it is easy to guess this? Whether or not a beginner can guess something doesn't make it not true or a "cop-out." – d_b Aug 03 '21 at 19:56
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    I cannot understand which parts of this are you quoting / paraphrasing Feynman and which parts are you talking yourself. Could you try to reformat it a bit to make this more clear. At least a paragraph break? – Brick Aug 03 '21 at 20:01
  • I've enclosed the quotes from Feynman lectures in long brackets @Brick –  Aug 03 '21 at 20:05
  • So might another version of this question be "How do we know that $(10^{is})^* = 10^{-is}$? I think, if I understand you, that your question is how we reason from the definitions that complex conjugation "passes through to the exponent" like that. – Brick Aug 03 '21 at 20:09

2 Answers2

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As you say, the only defining property of $i$ is $i^2=-1$. If you replace $i$ with $-i$, everything still works. Therefore, if $10^{is}=x+iy$ with $s,\,x,\,y\in\Bbb R$, $10^{-is}$ has to be $x-iy$, otherwise you could "tell apart" $i,\,-i$.

One can construct functions $f$ that don't satisfy $f(z^\ast)=f(z)^\ast$ but - without going into analytic functions, Cauchy-Riemann equations etc. - their definition must contain $i$, in order to break the symmetry. This is why, for example, that equation can be violated by $e^{iz}$ but not $e^z$.

J.G.
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  • @SeanE.Lake I've made an edit to address your comment. – J.G. Aug 03 '21 at 20:18
  • @Brick On point! –  Aug 03 '21 at 20:16
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    This doesn't answer the question - you need to bring in the fact that the exponential function is analytic. That's what allows you to "bring the complex conjugate in". – Sean Lake Aug 03 '21 at 20:15
  • @Brick the point is that complex conjugation can be viewed as an arithmetic operation $x+iy \mapsto x-iy$ or as a purely typographical one that replaces the symbol $i$ with the symbol $-i$. – jacob1729 Aug 03 '21 at 20:12
  • Pretty sure this begs the question. The complex conjugate operation is defined on the right hand side, but I think the question is why $(10^{is})^* = 10^{-is}$. That's not a definition. – Brick Aug 03 '21 at 20:11
  • That's Feynman's reason not mine. And that's what I don't find wholly satisfactory, (it is easy to find it "obvious" when you have prior acquaintance with complex analysis but may I remind you that the chapter is aimed at students who only know the algebra of real numbers and who have only just been convinced that certain numbers exist whose square is negative) –  Aug 03 '21 at 20:11
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    "That is about all we are going to say about it; of course, there is more than one root of the equation $x^2=−1$. Someone could write $i$, but another could say, “No, I prefer $−i$. My $i$ is minus your $i$.” It is just as good a solution, and since the only definition that $i$ has is that $i^2=−1$, it must be true that any equation we can write is equally true if the sign of $i$ is changed everywhere. This is called taking the complex conjugate. " <-Feynman, from the same section of that chapter. – jacob1729 Aug 03 '21 at 20:10
  • Another way to think of it is $i$ behaves as a unit vector, so changing to $-i$ is nothing more than a change of variables like $\hat{a}$ to $\hat{b}$ – Señor O Aug 03 '21 at 20:10
  • @jacob1729 But I guess the point is that you have to prove that the two are the same. (I don't dispute that they are, but they are not a-priori equivalent.) – Brick Aug 03 '21 at 20:13
  • @jacob1729 Maybe you should make your comment into an answer, it seems to exactly describe Feynman's reasoning behind it – doetoe Aug 03 '21 at 21:25
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    You are saying that if $10^{is}$ is well-defined, then it must enjoy certain properties. This is a very tenuous argument: it depends (1) on what the actual (unstated) definition of $10^{is}$ is and (2) on how that definition uses properties of $i$. I am very sympathetic to hand-waving arguments that help applied mathematicians understand important concepts, but I don't think your explanation is adequate. – Rob Arthan Aug 03 '21 at 22:24
  • @RobArthan The purpose of my answer is to explain what Feynman is saying, not to prove he's being rigorous. I'd have to re-read what the OP did to know whether he addresses your concerns earlier, e.g. "here's how to define $e^{it}$ with $t\in\Bbb R$". – J.G. Aug 04 '21 at 06:45
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Feynman in that whole section is using results that he knows to be true and trying to provide some intuition, but you've picked one of several things in that chapter that are not mathematically rigorous. Even just with the equations you've shown, a mathematician would want to prove that $10^{r+is} = 10^r 10^{is}$. It's true, but you cannot assume it's true just from the definition $i^2 = -1$. (For other structures, like matrix exponentiation, it's not generally true, just for example.)

Similarly, he's assuming that $(10^{is})^* = 10^{-is}$. Again, that's definitely true but it also does not immediately follow from the definitions. It should be proved if you want to be mathematically rigorous.

I don't know that this is a "cop out" because I also don't think he's claiming rigor here at all. He's trying to give some quick intuition and, like other teachers, he's approaching it with foresight to the answers in a way that he thinks is useful to the student. I think there's no more or less to it than that.

Brick
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    In physics we call this "hand waving", and it's really common. – Sean Lake Aug 03 '21 at 20:28
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    Rigour is not the point here. Plausibility is. Throughout the chapter he's being very ingenuous and resourceful with his exposition. Look for example how he treats logarithms - you only need to know basic arithmetic to follow his explanation about how to compute logarithms by hand! However this one( complex powers) explanation stands out as not so Feynmanesque –  Aug 03 '21 at 20:31
  • Another fact he assumed: the complex numbers are closed under algebraic operations and exponentiation. I mean, we went from integers to rationals using division, rationals to real numbers with rational exponents applied to positive numbers, and reals to complex with negative bases. – Sean Lake Aug 03 '21 at 20:36
  • @AdNKhan If that's the case, I don't really understand your objection. Both the left and right-hand sides of $e^{is}=x+iy$ can be viewed loosely as functions of $i$; in that sense, it is plausible that substituting $i\leftrightarrow -i$ on both sides preserves the equality. After all, $f(x,y)=e^{xy} \implies f(-x,y)=e^{-xy}$. How this is demonstrated rigorously depends on how you decide to formally construct the complex numbers, but in several approaches (e.g. defining $\mathbb C$ to be the algebraic closure of $\mathbb R$) the rigorous explanation parallels Feynman's heuristic one. – J. Murray Aug 03 '21 at 20:43
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    I'm in agreement with @J.Murray on his last point. It's certainly plausible a priori that this would be true. (And if fact it is true!) If you take it on faith / hand-waive / assume-it-for-now-and-prove-it-later, then the rest of his explanation is trying as quickly as possible to get back to regular arithmetic where he thinks there's more intuition. – Brick Aug 03 '21 at 20:50
  • @JMurray I understand your point but please consider this: What makes it implausible that e^is =e^(-is) =x+i*y ? It's easy to rule it aside as nonsense since one is the reciprocal of another but since we are stepping out of the real domain into the complex /imaginary domain, we're not entitled to assume that the reciprocals must be different numbers as they are in the case of real numbers ( Again lose any preconceived notions about complex algebra and try to approach it from a fresh background like Feynman does in that chapter) – Ad N Khan Aug 04 '21 at 11:58
  • Also can you please provide some resources/books that go into the more abstract aspects like how we define a complex number? What is a number anyway? Why do we define certain properties like for example multiplying complex numbers we multiply the real parts and subtract the product of the imaginary parts o get the real part of the product of the given complex numbers, and so on... What guides our definitions? – Ad N Khan Aug 04 '21 at 12:04
  • @AdNKhan If you're putting yourself back at the start as if you don't the answers, then you have to accept that more than one thing might seem plausible to you until you start proving or disproving them and getting a more complete picture. You're trying to make something out of this chapter that it isn't and was never meant to be. Any complex analysis book should cover the topics that you mentioned. I bet most of them have good answers here on math.SE as well. – Brick Aug 04 '21 at 13:22
  • @Brick If I must prove those assumptions rigorously before I can sift through equally reasonable /plausible assumptions, tell me what is the point of that whole discussion? It is like saying "OK, such and such notions appeal to intuition, good! As for the less obvious ones, since we know from formal considerations that they follow, they must be true. So let's claim that they are plausible just because rigorous approaches back us up." – Ad N Khan Aug 04 '21 at 14:04