# 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:

\(\ds \map \arg {z^{k + 1} }\) | \(=\) | \(\ds \map \arg {z z^k}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map \arg z + \map \arg {z^k}\) | Argument of Product equals Sum of Arguments | |||||||||||

\(\ds \) | \(=\) | \(\ds \map \arg z + k \map \arg z\) | by our induction hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {k + 1} \map \arg z\) |

$\blacksquare$