The principal argument for any negative real number without an imaginary part is always equal to π, while for positive real numbers, the argument is 0. The notation Arg(z) represents the principal value, which is confined to the interval (-π, π]. The argument for negative real numbers can be expressed as π + 2πk, where k is any integer, and for positive real numbers as 0 + 2πk. For purely imaginary numbers, the arguments are π/2 and -π/2, respectively.
PREREQUISITESStudents of mathematics, particularly those studying complex analysis, educators teaching complex number concepts, and anyone seeking to clarify the distinctions between Arg(z) and arg(z).
charmedbeauty said:Homework Statement
Its not really a problem I was just wondering if the argument for any negative real number but no imaginary part was always = to pi?
ie -1, -2,-3, -0.65... is the arg(z)=pi for all these cases
if so I am guessing for positive real numbers with no imaginary part then Arg(z)=0
and likewise for imaginary numbers with no real parts then Arg(z)= pi/2 and -pi/2 respectively.
Thanks
scurty said:You are using Arg(z) and arg(z) too loosely. Arg(z) for negative real numbers is \pi, and for positive real numbers, 0. arg(z) for negative real numbers is \pi + 2\pi k,\ k \in \mathbb{Z}, and for positive real numbers, 0 + 2 \pi k,\ k \in \mathbb{Z}. Remember that \text{Arg}(z) \in (-\pi,\pi]. You are correct for the imaginary numbers.
But, yes, the principal argument is always those four values you listed. The argument has an infinite number of values.