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There is something I fail to understand involving Leonhard Euler's identity:

It is well known that $(e^{2π})^i = 1$.

That means $\sqrt[i]{1} = e^{2π} ≈ 535.49 $.

But there's a rule that states $ \sqrt[a]{1} = 1$ , that means $\sqrt[i]{1} = 1$.

Applying this rule here: if $ a = b, a = c $ it means $ b = c $.

That means $ 535.49 = 1 $?

Martin Sleziak
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TreasureGhost
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  • You must be careful when taking powers of complex numbers. For example what is your definition of $\sqrt[i]{x}$ ? – Delta-u Apr 25 '18 at 13:13
  • Taking $i^{th}$ roots is really tricky. Search this site for "complex logarithm". Here's one place to start: https://math.stackexchange.com/questions/578477/complex-logarithm – Ethan Bolker Apr 25 '18 at 13:16
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    "But there's a rule that states $\sqrt[a] 1 = 1$". Well, when working in complex numbers, $\sqrt[3]1 = -\frac 12 + i\frac{\sqrt 3}{2}$ (one of the values anyway, look up the "cube roots of unity"). And that's definitely not $1$. If you can observe such "weird" behaviour even with an integer in place of $a$, why would you expect any less "weird" behaviour with an imaginary or complex number in that place? (Note that I'm not saying there's any paradox here - just that your assumption that $\sqrt[a] 1 = 1$ is completely mistaken when dealing with complex numbers. – Deepak Apr 25 '18 at 13:18
  • This is another example of essentially this question. – Xander Henderson Apr 25 '18 at 13:39

2 Answers2

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The only way to define $1^i$ is $=e^{i\ln 1}$ but $\ln 1$ does not have a single value over the complex numbers, so $$\ln 1=0, \pm 2\pi i, \pm 4\pi i, \ldots$$ thus $1^i$ is not a single value, but a set of possible values.

Rene Schipperus
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The rule

$$(a^p)^q=a^{pq}$$ doesn't hold for complex exponents, nor

$$(a^p)^{1/p}=a.$$

You just found a counterexample.