<s>[INST] Is the modulus of i^n 1 for all n? As the tile says, is $\left | i^{n} \right | = 1$ for all real values of n?
 [/INST] We note that $|i| = 1$: $|i| = |0+1i| = \sqrt{0^2+1^2} = \sqrt{1} = 1$.
Suppose that $|i^n| = 1$. Then, $|i^{n+1}| = |i^ni| = |i^n||i| = 1\cdot 1 = 1$. By induction, $|i^n| = 1$ for all positive integers $n$.

Another method:
In general, $|z|^2 = z\overline{z}$.
$$|i^n|^2 = (i^n)\overline{(i^n)} = \underbrace{i\ \overline{i} \cdots i\ \overline{i}}_{n \textrm{ times}} = 1\cdot 1 \cdots 1 = 1.$$

A third method:
$$z \cong \begin{pmatrix} a & -b \\ b & a\end{pmatrix}$$ and $$|z|^2 = \det \begin{pmatrix} a & -b \\ b & a\end{pmatrix}.$$
So $$|i^n|^2 = \det \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^n = \left[\det \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\right]^n = 1.$$
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