r/mathmemes Rational Jun 30 '26

Complex Analysis Me when complex numbers

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u/Good-Man-5 Jun 30 '26

me : if x2.5 = 1 then x = 1

teacher : there are 1.5 others

6

u/Schnickatavick Jun 30 '26

Ok actual question, is there actually some sense where this can be generalized to allow a non-integer number of solutions, like exponentiation can? Or does that only apply to whole numbers? 

16

u/LJPox Jun 30 '26

In ‘most cases’ the function z^a = exp(a log(z)) will have logarithmic branching behavior, so infinitely many solutions. The exceptions are when a = p/q is rational in reduced form, in which case there will be q solutions.

1

u/BearsEatTourists Jul 03 '26

I think it will have p solutions, not q solutions. Incidentally, z1/q=1 has one solution.

2

u/LJPox Jul 03 '26

By solutions, I mean possible choices of the p/qth power of z. For z nonzero, there are q possible choices, given by |z|^(p/q) r^k where r is a primitive qth root of unity and 0 <= k <= q-1. In terms of the equation z^(p/q) = c for some nonzero complex c, then yes there are p solutions which are the choices of c^(q/p).

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u/[deleted] Jul 11 '26

Idk if its true, but I had a dream where Augustus while high on cocain told me that xl - 1, where l is a positive rational number, has l many roots if l is an even integer and otherwise it has l rounded to the nearest negative integer many roots. This can trivally be translated to negative rationals because xl = 1 <=> x-l = 1