Argument of x to the n Equals n Times The Argument

From ProofWiki
Jump to navigation Jump to search


Let $z$ be a complex number.


$\forall n \in \N_{>0}: \map \arg {z^n} = n \map \arg z$


For $n = 1$

$\map \arg {z^1} = 1 \cdot \map \arg z$

Assuming the result is true for $n = k$, we have:

\(\ds \map \arg {z^{k + 1} }\) \(=\) \(\ds \map \arg {z z^k}\)
\(\ds \) \(=\) \(\ds \map \arg z + \map \arg {z^k}\) Argument of Product equals Sum of Arguments
\(\ds \) \(=\) \(\ds \map \arg z + k \map \arg z\) by our induction hypothesis
\(\ds \) \(=\) \(\ds \paren {k + 1} \map \arg z\)