Argument of x to the n Equals n Times The Argument

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Theorem

Let $z$ be a complex number.

Then:

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


Proof

For $n = 1$

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

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

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

$\blacksquare$