Argument of Negative Real Number is Pi
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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:
- $\arg \paren {-x} = \pi$
$\blacksquare$