Asymptotic series of a matrix-valued function. Consider the following matrix $$f(\lambda)=\left( \frac{\lambda-1}{\lambda + 1} \right)^{\nu \sigma_3} \ \ \ \lambda \in \mathbb{C} \setminus [-1,1]$$ where $\sigma_3=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ is the third pauli matrix and $\nu $ is a negative real number. Clearly $f(\lambda) \to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $ as $\lambda \to \infty$. But my question is, what is the asymptotic series of $f(\lambda)$ as $\lambda \to \infty$ ? 
Any help is appreciated, Thanks.

Set $a = (\lambda - 1)/(\lambda + 1)$, so that $a \to 1$ as $\lambda \to \infty$. Then
$$
\begin{align}
a^{-\nu\sigma_3} = \exp(-\nu\ln(a)\sigma_3) & = \sum_{k=0}^\infty{\frac{1}{k!}X^k},\quad X\equiv-\nu\ln(a)\sigma_3\\
& = I\left(\sum_{k=0}^\infty{\frac{\left(\nu\ln(a)\right)^{2k}}{(2k)!}}\right) - \sigma_3\left(\sum_{k=0}^\infty{\frac{\left(\nu\ln(a)\right)^{2k+1}}{(2k+1)!}}\right) \\
& = I\left(1 + \mathcal{O}(\ln^2(a))\right) - \sigma_3\left(\nu\ln(a) + \mathcal{O}(\ln^3(a))\right) \\
& \to I\quad (a\to1)
\end{align}
$$
where we've used the fact that $\sigma_3^{2k} = I$ and $\sigma_3^{2k + 1} = \sigma_3$.
The identity in the first line equating a matrix exponential to a power series is actually the definition of the matrix exponential, and can be thought of as an analytic continuation of the power series of $e^x$ for $x \in \Bbb{C}$.
