# Argument of x to the n Equals n Times The Argument

## 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 $\ds$ $=$ $\ds \paren {k + 1} \map \arg z$

$\blacksquare$