Why does (-1)^(-6/7) give me an error in my calculator?

aliveandwell

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Any comments are appreciated. Thank you
 

Subhotosh Khan

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[h=2]Why does (-1)^(-6/7) give me an error in my calculator?[/h]
Any comments are appreciated. Thank you
Because (-1)^(1/2) is an "imaginary" number. Most of the elementary calculators are NOT designed to handle complex numbers.
 

Dr.Peterson

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Why does (-1)^(-6/7) give me an error in my calculator?

Any comments are appreciated. Thank you
Although the result of (-1)^(-6/7) is not imaginary, so this shouldn't cause trouble ideally, calculators commonly use logarithms to calculate powers, and as a result they only accept positive bases. Some calculators evidently allow for special cases, such as (-1) to integer powers or to fractions with odd denominators, but others just apply the same method in all cases. (The calculator app on my computer has no trouble with your example, nor does one app I use on my phone, but another does.)
 

Dr.Peterson

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Interesting. Google says it's -0.900968868 - 0.433883739 i, while all my calculators that give an answer say 1. And, of course, both are right, in the sense that they are among the 7 7th roots of (-1)^6 = 1.

I've noticed before that some calculators choose oddly like that. I recall that Wolfram Alpha is among them; here is confirmation: https://www.wolframalpha.com/input/?i=(-1)^(-6/7) Note that it eventually gets around to showing all 7 roots.

I suppose I lied when I said it isn't imaginary; I should perhaps have said it doesn't have to be imaginary.
 
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Subhotosh Khan

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Interesting. Google says it's -0.900968868 - 0.433883739 i, while all my calculators that give an answer say 1. And, of course, both are right, in the sense that they are among the 7 7th roots of (-1)^6 = 1.

I've noticed before that some calculators choose oddly like that. I recall that Wolfram Alpha is among them; here is confirmation: https://www.wolframalpha.com/input/?i=(-1)^(-6/7) Note that it eventually gets around to showing all 7 roots.

I suppose I lied when I said it isn't imaginary; I should perhaps have said it doesn't have to be imaginary.
The primary root of the equation is "complex".
 

Dr.Peterson

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The primary root of the equation is "complex".
How are you defining "primary root"? Wolfram Alpha called their answer the "principal root" but I'm not sure what definition they are using there, as it does not fit this definition.

In fact, all the roots are complex, including 1. I suppose you mean "non-real".
 
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