Question

how come e^jπ=-1 ?????????

Answer 1

depends on what you think
\[e^x\] is

Answer 2

if you are using
\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] you get the answer by substitution

Answer 3

no its fixed...as euler pointed out

Answer 4

well if it ain't broke, don't fixed it

Answer 5

\[e=\sum_{k=0}^{\infty}\frac{1}{k!}\]