Why matlab return me a complex number

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David
David el 27 de Sept. de 2013
Respondida: Edu Santos el 21 de Mayo de 2019
Hi, Why >>(-27)^(1/3) return me a complex number rather than -3??? I am curious about this. Please help.

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Andreas Goser
Andreas Goser el 27 de Sept. de 2013
From the documentation
doc power
"Note that for a negative value X and a non-integer value Y, if the abs(Y) is less than one, the power function returns the complex roots. To obtain the remaining real roots, use the nthroot function."
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David
David el 27 de Sept. de 2013
Editada: David el 27 de Sept. de 2013
Thanks very much. I also found an useful answer here: http://www.mathworks.com/support/solutions/en/data/1-15M1N/?solution=1-15M1N

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Más respuestas (2)

Edu Santos
Edu Santos el 21 de Mayo de 2019
If instead you do:
(abs(-27))^(1/3)
you may have the good absolute value with the bad sign.
Storing the sign in a term, for example like this:
-27/abs(-27)
we can arrive to the expected answer by doing:
-27/abs(-27)*(abs(-27))^(1/3)
=-3
A simpler way is to just do
nthroot(-27,3)
=-3

Azzi Abdelmalek
Azzi Abdelmalek el 27 de Sept. de 2013
% By definition a^n=exp(n*log(a))
% also if a is a positive real number, log(-a)=log(a)+j*pi
a=-27;
n=1/3
out1=exp(n*log(a))
%the same as
out2=a^n

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