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e^(ix)=cosx+isinx

So, e^(i*i)=cosi+isini

Or e^-1=cosi+isini

Or 1/e + 0*i= cosi+isini

So, cosi=1/e and sini=0

But that's not the value of sin(i) that I found on the internet. These values are not even satisfying cos^2(x)+sin^2(x)=1.

What did I miss?

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# B What's wrong with this proof of sin(i)=0?

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