A similar post here but no answers. http://www.mathworks.com/matlabcentral/newsreader/view_thread/86495
Why matlab return me a complex number
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Hi, Why >>(-27)^(1/3) return me a complex number rather than -3??? I am curious about this. Please help.
採用された回答
Andreas Goser
2013 年 9 月 27 日
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."
その他の回答 (2 件)
Edu Santos
2019 年 5 月 21 日
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 件のコメント
Azzi Abdelmalek
2013 年 9 月 27 日
% 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 件のコメント
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