Is $(-1)^{\varepsilon}=1+i\pi \varepsilon$?

Question

Let $\varepsilon$ be defined as in the dual numbers. Using the standard matrix representation. I tried $(-1)^{\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right)}$ and got the following result: $\left( \begin{array}{cc} 1 & i \pi \\ 0 & 1 \\ \end{array} \right)$. But complex numbers should have different representations in matrix form, so I onder if my interpretation is correct.

Answer 1

No, $$ (-1)^\epsilon = (e^{i\pi})^{\epsilon} = e^{i\pi\epsilon} = \cos(\pi\epsilon) + i \sin(\pi\epsilon)$$ or, at the very least this is one of the ways that $(-1)^\epsilon$ can be defined. You can get other results using $-1 = e^{i\pi+2n\pi}$, $n\in\mathbb{N}\setminus\{0\}$; the function $z^\epsilon$ is a so-called a multi-valued function on a complex plane and can be defined in amny ways.

Answer 2

As you say, $f(\epsilon)=f’(0)\epsilon$ for any differentiable function $f$ defined on a real neighborhood of $0$, when $\epsilon^2=0$. To compute $(-1)^\epsilon$ you need to determine $f$. The matrix expression doesn’t make anything easier here.

Generally, $(-1)^x=e^{x\log(-1)}$ and $\log(-1)=i(2k+1)\pi$ is multivalued. So there is a choice $f_k(x)=e^{xi(2k+1)\pi}$ of function here for each integer. Making a choice here is called choosing a branch cut of $\log$. We have $f_k’(0)=(2k+1)i\pi$, so $f_k(\epsilon)=(2k+1)i\pi\epsilon$. Choosing the branch cut $k=0$ recovers your calculation, but it is one of infinitely many correct answers, as almost always happens with complex exponentials.