In every case I've ever seen, Euler's Identity is written as
eiπ+1=0
with the "positive" √−1. However, my understanding is that both i and −i are valid for √−1.
Does this mean that
e−iπ+1=0
is also a valid identity?
Answer
Yes.
e−iπ=cos(−π)+isin(−π)=−1
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