It works because a negative number has a unique real 5th root. Same thing if you raise it to the reciprocal of any odd integer. The exact process in which Excel casts the variant 0.2 to the reciprocal of the integer 5 is murky to me, but it seems to work in general.what algorithm is Excel using for taking x^y
assume x<0 and y between 0 and 1...is it then a bit strange excel gives output when y=0.2 (it should be a imaginary number, Mathematica get it right, but why is Excel giving error for all except y=0.2 ?) I would expect also error here?
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likely very stupid question, but I was just wondering if Excel could do imaginary numbers, and it looks like it can do some, but with wrong answer. Am I imagining this?
It does work in Excel, although I think it would take me a little time to get comfortable with the notation.As there are many solutions, we should select one. I used the following formula for LibreOffice Calc (on Linux) but it should work with Excel. (IM is for complex functions and a^b is evaluated as exp(log(a)*b))