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since $e^π= 23.14...$ and this to the power of $i = -1$ , would $23.05...$ equal $-.9995$ ? would $23.27...$ equal $-1.003$ ? in other words, does every positive real number generate a different value when raised to the $i$ power or are all $n^i$ equal to $-1$?

I've seen several videos about the Euler identity but but no one addresses the obvious non-mathematician's question. If there is a unique point on a curve in the complex plane that is $-1$ ... and that point corresponds to $e^π$ ... then this is an important fact. If every number generates that same $-1$, then Euler's identity appears trivial.

Hasan Heydari
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Bantokfomoki
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  • Welcome to MSE. Please use MathJax. – José Carlos Santos Oct 02 '17 at 08:22
  • @Bantokfomoki related with your question (I guess), here you have a more generic visualization of the pattern you are talking about: https://math.stackexchange.com/questions/1436167/is-this-a-valid-visualization-of-eulers-identity-as-a-more-generic-pattern – iadvd Oct 02 '17 at 08:29

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Complex exponentials with positive real bases have a very simple and unambiguous definition: $$ x^z = e^{z\cdot \ln x} $$ That means that if you change the value of $x$ slightly, you do change the value of the resulting exponent slightly as well. However, it doesn't go the way you think: Any (positive, real) number raised to a purely imaginary number has absolute value $1$ (because by the deifnition above, it corresponds to $e$ raised to some other purely imaginary number). What changes is its argument (i.e. angle in the complex plane). So, for instance, $$ 23.05^i \approx -0.999992 + 0.004i $$ is above $-1$ along the unit circle, while $$ 23.27^i \approx -0.99998 -0.0056i $$ is below.

Arthur
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