We know that negative number to the power of any

wingtipti

wingtipti

Answered question

2022-01-23

We know that negative number to the power of any integers or some fractions will always have a solution. Is it possible for us to solve (2)13 or (2)e, by modifying/extending our current number system?

Answer & Explanation

sineurosi0f

sineurosi0f

Beginner2022-01-24Added 14 answers

Consider the value of the "number" (2)e; this is equivalent to considering
x=(2)e
lnx=eln(2)
lnxe=ln(2)
elnxe=2
12elnxe=1
elnxe+ln12=1
Given the equation ea+bi=ea(cosb+isinb), we only need to find x such that lnxe=iπ+2niπln12 or lnx=eiπ+2iπ+eln2 which is
x=eeiπ+2iπ+eln2=2e(cos(eπ+2π)+isin(eπ+2π)
which is an infinite set of unique complex values in a circle of radius 2e

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