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}}\right]\left[e^{\ln{(x^{-i})}}\right]\\=\left[e^{i\ln{(x)}}\right]\left[e^{-i\ln{(x)}}\right]\\=\left[\cos{(\ln{(x)})} i\sin{(\ln{(x)})}\right]\left[\cos{(-\ln{(x)})} i\sin{(-\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})} i\sin{(\ln{(x)})}\right]\left[\cos{(\ln{(x)})} - i\sin{(\ln{(x)})}\right]\\=\left[\cos{(\ln{(x)})}\right]^2 - i\cos{(\ln{(x)})}\sin{(\ln{(x)})} i\cos{(\ln{(x)})}\sin{(\ln{(x)})}-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2-i^2\left[\sin{(\ln{(x)})}\right]^2\\=\left[\cos{(\ln{(x)})}\right]^2 \left[\sin{(\ln{(x)})}\right]^2\\=1\end{array}\u0026bg\u003dffffff\u0026fg\u003d000000\u0026s\u003d0)