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Can someone please explain how to get to the above conclusion for T from the equation provided? I understand:

$e^{iwT} = 1$

Can be written as:

$\cos(wT) + i\sin(wT) = 1$

But I’m at a loss as how to proceed further. Though I know:

$T = \frac{2\pi}{w} $

It’s been a while since I’ve done this kind of math so if there’s a resource anyone can recommend for a refresh I’m open to suggestions. Also if there are better tags for this feel free to let me know.

Martin Sleziak
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IGotsQuestions
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  • $1$ can also be written as $1+0i$, so if you separate your second equation into real and imaginary parts you get two equations. Can you see what they are, and go from there? – JonathanZ Mar 09 '22 at 03:58
  • I think so, wouldn’t that mean $T = \frac{k2\pi}{w}$ where k = 0,1,2,3… Or am I still missing the mark completely? – IGotsQuestions Mar 09 '22 at 11:17
  • Yes, that is the right answer (although you need to include negative values for $k$ as well), but you'd need to provide more details on how you got that in order for us to tell you if you solved it correctly or just got lucky with some guessing. – JonathanZ Mar 09 '22 at 14:48
  • Also, to be complete, I will point out that the claim in your title isn't fully correct, as it's missing the $k$ that you (correctly) included in your comment. – JonathanZ Mar 09 '22 at 17:39
  • Sorry, here’s what I did: $ cos(wT) + isin(wT) = 1$ $ cos(wT) + isin(wT) = 1 + i0$ Separating the real and imaginary: $ cos(wT) = 1 $ $ isin(wT) = i0 $ , here I take the imaginary part of the equation to just be equal to zero so I ignore it $arccos(cos(wT)) = arccos(1) $ $ wT = k(2\pi) $ , for all integer values of k right? – IGotsQuestions Mar 09 '22 at 23:01
  • I'd say that that looks pretty good. The two quibbles I would make are that 1) because arccos() isn't always a single-valued inverse of cos(), writing $arccos(cos(wT))$ raises some questions, and it's better to go from $cos(wT) = 1$ to $wt = arccos(1)$, and then say that $wT = k2\pi$, and 2) you should at least mention that the angles $k2\pi$ do in fact have a a sin() of $0$. – JonathanZ Mar 09 '22 at 23:39
  • Gotcha, thanks for helping me out! – IGotsQuestions Mar 10 '22 at 02:03

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