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

Posted 11 years ago

Given i^(2/7), I get the following result:

i^(2/7)

Which doesn't look correct to me. So, I entered "7th root of -1", which gave me this:

7th root of -1

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)":

equality

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

equality

Imaginary power

i^(2/7)

7th root of -1

7th root of -1

POSTED BY: Joshua Gerrard
9 Replies
Posted 11 years ago

Ah right, that explains it. Thanks again!

POSTED BY: Joshua Gerrard
Posted 11 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 BY: Greg Hurst
Posted 11 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 BY: Joshua Gerrard
Posted 11 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 BY: Greg Hurst
Posted 11 years ago

Nevermind, found out how to run it.

POSTED BY: Joshua Gerrard
Posted 11 years ago
POSTED BY: Joshua Gerrard
Posted 11 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 BY: Joshua Gerrard

i agree something is amiss. Reported it as needing investigation.

POSTED BY: Daniel Lichtblau

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 BY: S M Blinder
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