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

\(\ds \)

\(=\)

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

$\blacksquare$