A similar post here but no answers. http://www.mathworks.com/matlabcentral/newsreader/view_thread/86495
Why matlab return me a complex number
14 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
David
le 27 Sep 2013
Réponse apportée : Edu Santos
le 21 Mai 2019
Hi, Why >>(-27)^(1/3) return me a complex number rather than -3??? I am curious about this. Please help.
Réponse acceptée
Andreas Goser
le 27 Sep 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."
Plus de réponses (2)
Edu Santos
le 21 Mai 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
0 commentaires
Azzi Abdelmalek
le 27 Sep 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
0 commentaires
Voir également
Catégories
En savoir plus sur Special Functions dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!