# 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?

• Yes. Do you know how to compute complex exponentials using polar coordinates?
– anon
Apr 24, 2014 at 19:06
• Hint: $|a^n|=|a|^n$, for all $n\in\mathbb{N}$. Apr 24, 2014 at 19:06
• @vadim123 Is that for all natural values of n or for all real values of n though? Apr 24, 2014 at 19:11
• @LazySloth13 It is definitely true for $a \in \mathbb{C}$ and $n \in \mathbb{N}$. It can be extended with care to other chosen values of $n$, if so desired. This extensibility has less to do with the value of the exponent, and more to do with the definition of $|\cdot|$, as exponentiation for positive reals can be defined for all real-valued exponents, i.e. $a^b$ exists uniquely for $a > 0$, $b \in \mathbb{R}$. Apr 24, 2014 at 19:18

Since $|i^n|=|i|^n$, and $|i|=1$, then for all $n\in\mathbb R$ $|i^n|=|i|^n=1$ for the simple reason that $1$ raised to any real exponent is still $1$.

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 $$|i^n|^2 = \det \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}^n = \left[\det \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\right]^n = 1.$$

• These proofs establish the rule for positive integer powers. However, the key takeaway is that we can write $|z^n| = |z|^n$ in general. Noting that $|z|$ is just a real number, we can use the formal construction of exponentiation to define $|z^x| = |z|^x$. Apr 24, 2014 at 19:34
• for the sake of pointing it out to curious readers, Arkamis's third method is just my argument about rotations in the language of matrix representations. Apr 24, 2014 at 19:35

Here's a less algebraic, more geometric explanation: $i$ denotes the vector $(0,1)$ in the complex plane, and multiplication by $i$ just rotates points in the plane 90 degrees counterclockwise. So $i^n$ is just a rotation of $(0,1)$ by 90 degrees counterclockwise $n-1$ times, i.e. a rotation of $(0,1)$ by $90(n-1)$ degrees. But $(0,1)$ has unit length, and rotations preserve length.

(This explains why $|i^n|=1$ for $n\in\mathbb{N}$. I didn't see that you wanted $n$ to be real. You're going to have trouble defining $i^{\frac{1}{n}}$ uniquely.)