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Euler's identity

$ e^{j \pi} + 1 = 0, \,\! $

Euler's formula

$ e^{jx} = \cos x + j \sin x \,\! $

$ \cos x = \mathrm{Re}\{e^{jx}\} ={e^{jx} + e^{-jx} \over 2} $

$ \sin x = \mathrm{Im}\{e^{jx}\} ={e^{jx} - e^{-jx} \over 2i}. $

$ \cos(x) = {e^{-jx} + e^{jx} \over 2} $

$ \sin(x) = {e^{-jx} - e^{jx} \over 2j} $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva