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得 搞掂喇
reader.elsevier.com/reader/sd/pii/S0024379504000771?token=7FB40A ... 3D0F7A502A07AC3D9A41
refer to Theorem 3.1
$X=\begin{pmatrix} \frac{k^{1/n}(1+\rho)+2u}{2} & v\frac{k^{1/n}(1-\rho)+2u}{2}\\
\frac{k^{1/n}(1-\rho)-2u}{2v} & \frac{k^{1/n}(1+\rho)-2u}{2}\end{pmatrix}$
take $k^{1/n}=exp\frac{2\pi i}{n},~\rho=exp\frac{-4\pi i}{n}$
$k^{1/n}(1+\rho)=exp\frac{2\pi i}{n}+exp\frac{-2\pi i}{n}=2\cos\frac{2\pi}{n}$
$k^{1/n}(1-\rho)=exp\frac{2\pi i}{n}-exp\frac{-2\pi i}{n}=2i\sin\frac{2\pi}{n}$
$k^{2/n}(1-\rho)^2=-4\sin^2\frac{2\pi}{n}$
$m=v\frac{k^{1/n}(1-\rho)+2u}{2}$
$\frac{k^{1/n}(1-\rho)-2u}{2v}=\frac{-\sin^2\frac{2\pi}{n}-u^2}{m}$
$X=\begin{pmatrix} \cos\frac{2\pi}{n}+u & m\\
\frac{-1}{m}(\sin^2\frac{2\pi}{n}+u^2) & \cos\frac{2\pi}{n}-u\end{pmatrix}$
when u=0, $X=\begin{pmatrix} \cos\frac{2\pi}{n} & m\\
\frac{-1}{m}\sin^2\frac{2\pi}{n} & \cos\frac{2\pi}{n}\end{pmatrix}$
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