Argument of Negative Real Number is Pi

Theorem

Let $x \in \R_{>0}$ be a positive real number.

Then:

$\arg \paren {-x} = \pi$

where $\arg$ denotes the argument of a complex number.

Proof

We have that:

$-x = -x + 0 i$

and so:

\(\ds \cmod {-x}\)

\(=\)

\(\ds \sqrt {\paren {-x}^2 + 0^2}\)

Definition of Complex Modulus

\(\ds \)

\(=\)

\(\ds x\)

Hence:

\(\ds \cos \paren {\arg \paren {-x} }\)

\(=\)

\(\ds \dfrac {-x} x\)

Definition of Argument of Complex Number

\(\ds \)

\(=\)

\(\ds -1\)

\(\ds \leadsto \ \ \)

\(\ds \arg \paren {-x}\)

\(=\)

\(\ds \pi\)

Cosine of Multiple of Pi

\(\ds \sin \paren {\arg \paren {-x} }\)

\(=\)

\(\ds \dfrac 0 x\)

Definition of Argument of Complex Number

\(\ds \)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \arg \paren {-x}\)

\(=\)

\(\ds 0 \text { or } \pi\)

Sine of Multiple of Pi

Hence:

$\arg \paren {-x} = \pi$

$\blacksquare$