\(e^{(-1+2j)t}\)+\(e^{(-1-2j)t}\)
my attempt:
\(e^{-t}e^{j2t}\) + \(e^{-t}e^{-j2t}\)
\(e^{-t}(e^{j2t}\) + \(e^{-j2t})\)
euler's identity says...
cos(2t)=\(e^{j2t}\)+\(e^{-j2t}\)/2
so...
\(e^{(-1+2j)t}\)+\(e^{(-1-2j)t}\) = \(e^{-t}(2cos(2t))\)
it seems right, but it's wrong! I checked the proof by substituting numbers into the original equation and the final equation and they evaluate to different numbers.
Any help would be nice.
my attempt:
\(e^{-t}e^{j2t}\) + \(e^{-t}e^{-j2t}\)
\(e^{-t}(e^{j2t}\) + \(e^{-j2t})\)
euler's identity says...
cos(2t)=\(e^{j2t}\)+\(e^{-j2t}\)/2
so...
\(e^{(-1+2j)t}\)+\(e^{(-1-2j)t}\) = \(e^{-t}(2cos(2t))\)
it seems right, but it's wrong! I checked the proof by substituting numbers into the original equation and the final equation and they evaluate to different numbers.
Any help would be nice.
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