Is the number #sqrt(-16)# real, complex, pure imaginary, or nonreal complex?

1 Answer
George C.
Aug 14, 2016

It is "complex", "pure imaginary" and "non-real complex".

It is not a "real" number.

Explanation:

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#sqrt(-16) = sqrt(16)i = 4i#

This Complex number has no Real part, so is "pure imaginary".

Note that any Real number is also a Complex number - it just has a zero imaginary part. So if we want to be explicit that a given number is not a Real number, the most precise term we can use is "non-real Complex".

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Footnote

The terminology surrounding Real and Complex numbers is rather unfortunate. Complex numbers are just as much numbers as Real numbers are. They just behave a little differently in some ways. For example:

  • Real numbers have a natural ordering, while Complex number do not.
  • Complex numbers contain all of the zeros of polynomials with Complex coefficients. Real numbers do not contain all of the zeros of polynomials with Real coefficients.
  • The function #e^x# is one to one from #(-oo, oo)# onto #(0, oo)# so has a well defined inverse function #ln(x)# from #(0, oo)# to #(-oo, oo)#
  • The function #e^z# is many to one from #CC# onto #CC "\" { 0 }#. We can define a principal natural logarithm #ln z# from #CC "\" { 0 }# into #CC# satisfying #e^(ln z) = z#, but #ln(e^z) = z# does not always hold.
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