I've figured out the following solution (probably not that hard anyway):
n1, n2 = -27, 3
root = ( n1 ** (1/n2))
print(root)
(1.5000000000000004+2.598076211353316j)
Unless I'm making a very dumb mistake, it should print -3 instead.
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martineau
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FishyBusinessYT
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3Related: [Cubic root of the negative number on python](https://stackoverflow.com/q/1361740/3890632). – khelwood Nov 30 '21 at 21:54
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one of strangest things I have seen in Python so far, when you do `-27 ** (1/3)` result is correct – dankal444 Nov 30 '21 at 21:54
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1@dankal444 That's because you're doing the equivalent of `-(27**(1/3))`. – khelwood Nov 30 '21 at 21:54
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Ah, of course.. my bad – dankal444 Nov 30 '21 at 21:55
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In general, there are 3 solutions to `z**(1/3)` for some complex number `z`. Python returns only one of them. – Scriniary Nov 30 '21 at 21:56
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The easiest way would be to get the cube root of positive 27, and negate the result. – khelwood Nov 30 '21 at 21:58
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@khelwood, that actually works, thank you very much! – FishyBusinessYT Nov 30 '21 at 21:58
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I still don't get why that happens though – FishyBusinessYT Nov 30 '21 at 21:59
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@FishyBusinessYT it happens because Python computes the _ principal cube root_ of a number, and there are mathematical reasons to prefer that over the other two roots. – Alessandro Cosentino Nov 30 '21 at 22:12
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This has to do with mathematics of complex numbers. There are three possible roots, one is -3 but the other two are complex. Python just returns the first root it finds. See this link for all 3 roots: https://www.wolframalpha.com/input/?i=%28-27%29%5E%281%2F3%29 – Marick Manrho Nov 30 '21 at 21:57
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It's not the "first root it finds", but it's the _principal_ root, which is well-defined mathematically. – Alessandro Cosentino Nov 30 '21 at 22:14
1 Answers
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To find the real cubic root of a negative number, use:
number = -27
if number > 0:
real_root = number ** (1/3)
elif number < 0:
real_root = - (abs(number) ** (1/3))
elif number == 0:
real_root = 0
jezza_99
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