how do you use e^a+bi = e^(cos b + i sin b) to derive the equation eπi + 1 = 0
I think you probably mean:
e^(a+bi) =e^a e^bi = e^a (cosb+isinb)
then
e^iπ = e^(0+iπ) =(e^0)(cos pi+isinpi)
= 1(-1 + 0)
or
e^iπ = -1
e^(a+bi) =e^a e^bi = e^a (cosb+isinb)
then
e^iπ = e^(0+iπ) =(e^0)(cos pi+isinpi)
= 1(-1 + 0)
or
e^iπ = -1