Can the Pythagorean theorem be proved using imaginary numbers? The proof must avoid circular reasoning, of course.
I asked essentially the same question at MSE, but did not receive a definitive answer, so I thought I would ask here.
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Yes. Write a complex number $a+bi$ in polar form $ce^{it}$. Then $$a^2+b^2=(a+bi)(a-bi)=(ce^{it})(ce^{-it})=c^2.$$ This is the first theorem I prove in my complex analysis class (after defining complex multiplication via the polar form and checking that it agrees with the more traditional definition via $i^2=-1$ and distributivity).
Added. See the comments for more details what this proof is really based on, and why it is not circular.
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5$\begingroup$ According to my understanding, this proof has met with objections, here and here. I believe the gist of the objections is that, in this proof, distance is implicitly being defined using the Pythagroean theorem, thus the reasoning is circular. I'm not sure what to make of the objections. $\endgroup$– DanCommented 11 hours ago
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4$\begingroup$ @Dan No. In my class, I base complex arithmetic on Euclidean geometry where distance is given (it is not a defined quantity but part of the model). It is a genuine proof just as any proof in Euclidean geometry. The difference is that here complex arithmetic is based on Euclidean geometry, and then the Pythagorean theorem becomes a one-liner as in my post. I leave you to fill in the details (or come to my class). $\endgroup$ Commented 11 hours ago
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3$\begingroup$ @GHfromMO Do you have class notes on a website, so that we can see that approach fleshed out? $\endgroup$ Commented 9 hours ago
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1$\begingroup$ @GHfromMO The context that you have defined complex arithmetic through Euclidean geometry is absolutely essential, and it should not be left in the comments. Please include it in your answer body, because one could interpret your answer as being just like many other identical circular answers that I’ve seen. At any rate, I might still have lingering questions e.g. it seems as if your primitive notion of Euclidean distance is real-valued, and I’d have to think a little more to find others $\endgroup$– FShrikeCommented 8 hours ago
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1$\begingroup$ All right, thank you for the explanation. $\endgroup$ Commented 5 hours ago