Interesting Math Equation

[ccb056]

a few days ago I can accross the equation: e^(2pi*i)-1=0
therefore
e^(2pi*i)=1
ln(e^(2pi*i))=ln(1)
2pi*i=0

since 2 and pi are both constants, you divide bot out and are left with: i=0

but 0 is a real number
but i is an imaginary number
how can an imiganary number be the same as a real number?

[The_Man]

ur solving the equasion is wrong, the reason it works out is because 1^i = 1