The principle argument when no imaginary part

[charmedbeauty]

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

Homework Equations

The Attempt at a Solution

 

[scurty]

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

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.

 

[charmedbeauty]

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.

ok thanks just wanted to clarify. thankyou!