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This question already has an answer here:

$e^{i\theta}=\cos\theta +i\sin\theta$. By that statement, we conclude that e^iθ is a circle on the complex plane for real values of $\theta$ because the second side of the equation $\cos\theta +i\sin\theta$ can easily be thought of as a circle. So my question is if we didn't have the formula how can $e^{i\theta}$ be thought as a circle on the complex plane ( and genially how can we explain that a real number to an imaginary power corresponds to circle on the complex plane without using the Taylor expansion )

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marked as duplicate by Jyrki Lahtonen, Tyrone, Taroccoesbrocco, Shailesh, Adrian Keister Aug 1 '18 at 12:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The parenthetical remark at the end really approaches the point. How do you define a power of real number base and an imaginary exponent? Taylor expansion is one of giving it a meaning. We can, of course, also dictate that such a power should follow certain rules, may be a differential equation also. But the point is, to significant extent, about extending the real definitions in such a way that some properties suvrive. $\endgroup$ – Jyrki Lahtonen Aug 1 '18 at 8:16
  • $\begingroup$ @BillIconomou Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Sep 6 '18 at 20:23
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The exponential form $e^{i\theta}$ corresponds to a complex number $z$ such that $|z|=1$ and $Arg(z)=\theta$ therefore varing $\theta$ the expression represents all the complex numbers on the unit circle centered at the origin.

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  • $\begingroup$ Well isn't there a more intuitive approach to this, and why does |z|=1 $\endgroup$ – Bill Iconomou Aug 1 '18 at 7:22
  • $\begingroup$ Refer also to math.stackexchange.com/q/2660361/505767 $\endgroup$ – gimusi Aug 1 '18 at 7:40
  • $\begingroup$ @BillIconomou, $\lvert z \rvert = \lvert \mathrm{e}^{i\theta}\rvert = \lvert \cos \theta + i \sin \theta\rvert = \cos^2 \theta + \sin^2 \theta = 1$ $\endgroup$ – Babelfish Aug 1 '18 at 8:00
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Notice that if we calculate the magintude of $e^ {i \theta}$ (that is, finding how far away it is from the origin) we have

$$| e ^ {i \theta} | = |cos(\theta) + isin(\theta)| = cos^2(\theta) + sin^2(\theta) = 1$$

Now, the magnitude is independent of $\theta$, so when rotating we'll always be the same distance from the origin, just like a circle where every point on its boundary is the same distance from its center.

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  • $\begingroup$ Yeah but in order to calculate it you use the Euler formula $\endgroup$ – Bill Iconomou Aug 1 '18 at 19:30

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