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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?

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  • $\begingroup$ Look in to complex analysis, Euler's formula, and logarithms. $\endgroup$
    – Sean Lake
    Commented Aug 3, 2021 at 19:55
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    $\begingroup$ 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." $\endgroup$
    – d_b
    Commented Aug 3, 2021 at 19:56
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    $\begingroup$ 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? $\endgroup$
    – Brick
    Commented Aug 3, 2021 at 20:01
  • $\begingroup$ I've enclosed the quotes from Feynman lectures in long brackets @Brick $\endgroup$
    – Ad N Khan
    Commented Aug 3, 2021 at 20:05
  • $\begingroup$ 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. $\endgroup$
    – Brick
    Commented Aug 3, 2021 at 20:09

2 Answers 2

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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 $e^z$.

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  • $\begingroup$ 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 $\hat{b}$ $\endgroup$
    – Señor O
    Commented Aug 3, 2021 at 20:10
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    $\begingroup$ "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. $\endgroup$
    – jacob1729
    Commented Aug 3, 2021 at 20:10
  • $\begingroup$ 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) $\endgroup$
    – Ad N Khan
    Commented Aug 3, 2021 at 20:11
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    $\begingroup$ 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". $\endgroup$
    – Sean Lake
    Commented Aug 3, 2021 at 20:15
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    $\begingroup$ 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. $\endgroup$
    – Rob Arthan
    Commented Aug 3, 2021 at 22:24
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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.

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    $\begingroup$ In physics we call this "hand waving", and it's really common. $\endgroup$
    – Sean Lake
    Commented Aug 3, 2021 at 20:28
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    $\begingroup$ 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 $\endgroup$
    – Ad N Khan
    Commented Aug 3, 2021 at 20:31
  • $\begingroup$ 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. $\endgroup$
    – Sean Lake
    Commented Aug 3, 2021 at 20:36
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Commented Aug 3, 2021 at 20:43
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    $\begingroup$ 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. $\endgroup$
    – Brick
    Commented Aug 3, 2021 at 20:50

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