How does y = e^(x(1-i)) + e^(x(1+i))(adsbygoogle = window.adsbygoogle || []).push({});

work out to y = (e^x)sinx + (e^x)cosx???

Using Euler's identity I get,

y = (e^x)e^-ix + (e^x)e^ix

y = e^x(cosx - isinx + cosx + isinx)

y = e^x(2cosx)

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# Euler's identity

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