We know that
(−1)=cos(2n+1)π±isin(2n+1)π=e±(2n+1)πi,for n=1,2,3,⋯ (-1) = \cos (2n+1)\pi \pm i\sin (2n+1)\pi = e^{\pm (2n+1)\pi i}, \text{for $n=1,2,3,\cdots$}(−1)=cos(2n+1)π±isin(2n+1)π=e±(2n+1)πi,for n=1,2,3,⋯
Therefore,
(−1)2i=e(±(2n+1)πi)⋅2i=e∓2(2n+1)π,for n=1,2,3,⋯ (−1)2i=e∓mπ,for m=2,4,6,⋯ (-1)^{2i} = e^{(\pm(2n+1)\pi i) \cdot 2i}= e^{\mp 2(2n+1)\pi}, \text{for $n=1,2,3,\cdots$}\\ (-1)^{2i} = e^{\mp m\pi}, \text{for $m=2,4,6,\cdots$}\\(−1)2i=e(±(2n+1)πi)⋅2i=e∓2(2n+1)π,for n=1,2,3,⋯(−1)2i=e∓mπ,for m=2,4,6,⋯
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