Here's a question more accessible to the majority of MX2 students.
I have stated that they are non-zero real numbers.x,y,z are non-zero complex numbers?
I just solved it:I have stated that they are non-zero real numbers.
Well, there is something called "analytic continuation".However I believe that extends it to non-zero complex numbers as well.
Well yes, if I prove it to be true for complex numbers, then it must be true for the real numbers. But I was just pointing out that the question can be extended to non-zero complex numbers.Well, there is something called "analytic continuation".
I don't think there are non-zero real numbers x, y and z satisfying x + y + z = 1/x + 1/y + 1/z = 0. If there are such real numbers, what are they?I have stated that they are non-zero real numbers.
Yes I think there are no all three real solutions. This comes from the fact that, if we have a polynomial roots x, y, z with the restrictions in the question.There isn't any. You can prove it by rearranging the conditions to get it in terms of z;
.
Then form the difference function
which can be shown with some basic multivariate calculus to not have any zeros among the reals. (well, atleast that's what my quick working shows. I might be wrong.)
That's true. It also gives us that the only possible form of x,y and z are which after a quick substitution prove to be solutions.Yes I think there are no all three real solutions. This comes from the fact that, if we have a polynomial roots x, y, z with the restrictions in the question.
Then the polynomial equation will be:
Where it is a polynomial in alpha, and d is any constant, from this equation we can automatically deduce that there can only be 1 real root (and non-zero).
No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).Well yes, if I prove it to be true for complex numbers, then it must be true for the real numbers.
The reals are a subset of C, so if proven true for all C, then it is true for all R. The converse isn't necessarily the same.No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
The real numbers are apart of the complex numbers though, so proving true for all complex numbers would imply it's true for all real numbers. Like all real numbers can be expressed as a complex numberNo actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
or this.The reals are a subset of C, so if proven true for all C, then it is true for all R. The converse isn't necessarily the same.
you're thinking about analytic continuity, right?No actually I THINK it's the other way around... It's been a while since I've done complex analysis though (which is VERY beautiful).
So? I don't see how that is relevant here.Well, there is something called "analytic continuation".
Cbb checking haha but that is the correct idea.DONE:
I'm not bothered to post working out, basically let x = that root, then find x^6, x^4 and x^2 using binomial theorem, then manipulation then:
I better not of made a silly mistake :/
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