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 2013년 9월 27일

0 개 추천

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."

추가 답변 (2개)

Edu Santos
Edu Santos 2019년 5월 21일

1 개 추천

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 2013년 9월 27일

0 개 추천

% 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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질문:

2013년 9월 27일

답변:

2019년 5월 21일

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