ArcTan[(-2 + z^2)/(2 Sqrt[z^2 - 1])] == (Pi/2) (z/Sqrt[-1 + z^2]) Sqrt[1 - 1/z^2] (Sqrt[I/z] Sqrt[(-I) z] - Sqrt[-(I/z)] Sqrt[I z] + Sqrt[1/z^2] z - Sqrt[(-1 + z)/z] Sqrt[z/(-1 + z)] + Sqrt[1 + 1/z] Sqrt[z/(1 + z)]) - 2 (z/Sqrt[-1 + z^2]) Sqrt[1 - 1/z^2] ArcSin[1/z]

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</mo> <mi> z </mi> </mrow> <mo> - </mo> <mrow> <msqrt> <mfrac> <mrow> <mi> z </mi> <mo> - </mo> <mn> 1 </mn> </mrow> <mi> z </mi> </mfrac> </msqrt> <mo> &#8290; </mo> <msqrt> <mfrac> <mi> z </mi> <mrow> <mi> z </mi> <mo> - </mo> <mn> 1 </mn> </mrow> </mfrac> </msqrt> </mrow> <mo> + </mo> <mrow> <msqrt> <mfrac> <mi> &#8520; </mi> <mi> z </mi> </mfrac> </msqrt> <mo> &#8290; </mo> <msqrt> <mrow> <mrow> <mo> - </mo> <mi> &#8520; </mi> </mrow> <mo> &#8290; </mo> <mi> z </mi> </mrow> </msqrt> </mrow> <mo> - </mo> <mrow> <msqrt> <mrow> <mo> - </mo> <mfrac> <mi> &#8520; </mi> <mi> z </mi> </mfrac> </mrow> </msqrt> <mo> &#8290; </mo> <msqrt> <mrow> <mi> &#8520; </mi> <mo> &#8290; </mo> <mi> z </mi> </mrow> </msqrt> </mrow> <mo> + </mo> <mrow> <msqrt> <mrow> <mn> 1 </mn> <mo> + </mo> <mfrac> <mn> 1 </mn> <mi> z </mi> </mfrac> </mrow> </msqrt> <mo> &#8290; </mo> <msqrt> <mfrac> <mi> z </mi> <mrow> <mi> z </mi> <mo> + </mo> <mn> 1 </mn> </mrow> </mfrac> </msqrt> </mrow> </mrow> <mo> ) </mo> </mrow> </mrow> <mo> - </mo> <mrow> <mfrac> <mrow> <mn> 2 </mn> <mo> &#8290; 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2003-08-21