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# Is this a bug? i^(2/7) yields different results to 7th root of -1

Posted 9 years ago
 Given i^(2/7), I get the following result: Which doesn't look correct to me. So, I entered "7th root of -1", which gave me this: Which is what I expected. So I thought, perhaps these aren't equivalent and I've gone wrong somewhere? I entered "(7th root of -1) = i^(2/7)": WolframAlpha appears to be contradicting itself! Looking more closely, the two seem to be rotations of one another. Is this a bug, or am I missing something here? Links Equality Imaginary power 7th root of -1 9 Replies
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Posted 9 years ago
 Thank you for finding this. We've fixed this today, and it should make it to the live site in about a week.
Posted 9 years ago
 Glad to help! If you're able to tell me, what was causing the issue? Otherwise thank you for such a great tool.
Posted 9 years ago
 The numerator in the fractional exponent was being overlooked, so you were seeing the root plot for i^(1/7) instead of i^(2/7).
Posted 9 years ago
 Ah right, that explains it. Thanks again!
Posted 9 years ago
 i agree something is amiss. Reported it as needing investigation.
Posted 9 years ago
 Thank you for your help, it did strike me as odd and my more mathematically adept peers at work agreed. Hopefully this will lead to an improvement in WolframAlpha in some way or form!
Posted 9 years ago
 Noting that E^(I Pi(1+2n))=-1 for {n,0,6} Do[Print[Re[E^(I Pi (1 + 2 n)/7)], " ", Im[E^(I Pi (1 + 2 n)/7)]], {n, 0, 6}] Cos[\[Pi]/7] Sin[\[Pi]/7] Sin[\[Pi]/14] Cos[\[Pi]/14] -Sin[(3 \[Pi])/14] Cos[(3 \[Pi])/14] -1 0 -Sin[(3 \[Pi])/14] -Cos[(3 \[Pi])/14] Sin[\[Pi]/14] -Cos[\[Pi]/14] Cos[\[Pi]/7] -Sin[\[Pi]/7] 
Posted 9 years ago
 Is this some form of source code?Apologies if I'm missing something obvious, I am not particularly knowledgeable in the world of maths (this came from me experimenting with fractional powers of complex numbers to learn about them more deeply).
Posted 9 years ago
 Nevermind, found out how to run it.