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<math>
\sum_{k=0}^\infty{1 \over k!}B_n^k =\begin{pmatrix}
\sum_{k=0}^\infty{(-1)^k \over 2k!}(\alpha+2\pi n)^{2k} & -\sum_{k=0}^\infty{(-1)^k \over (2k+1)!}(\alpha+2\pi n)^{2k+1} \\
\sum_{k=0}^\infty{(-1)^k \over (2k+1)!}(\alpha+2\pi n)^{2k+1} & \sum_{k=0}^\infty{(-1)^k \over 2k!}(\alpha+2\pi n)^{2k} \\
\end{pmatrix} =\begin{pmatrix}
\cos(\alpha) & -\sin(\alpha) \\
\sin(\alpha) & \cos(\alpha) \\
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Suppose that ''A'' and ''B'' commute, meaning that ''AB'' = ''BA''. Then
: <math>\log{(AB)} = \log{(A)}+\log{(B)}</math>
if and only if <math>\operatorname{arg}(\mu_j) + \operatorname{arg}(\nu_j) \in (- \pi, \pi]</math>, where <math>\mu_j</math> is an [[eigenvalue]] of <math>A</math> and <math>\nu_j</math> is the corresponding [[eigenvalue]] of <math>B</math>.<ref>{{cite journal |last1=APRAHAMIAN |first1=MARY |last2=HIGHAM |first2=NICHOLAS J. |title=The Matrix Unwinding Function, with an Application to Computing the Matrix Exponential |journal=SIAM Journal on Matrix Analysis and Applications |year=2014 |volume=35 |issue=1 |page=97 |doi=10.1137/130920137
: <math> \log{(A^{-1})} = -\log{(A)}.</math>
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