Question
1i*ln(1/[e^pi]^1i) simplify ==> -pi and 1i*ln(1/[e^pi]^1i) = -pi simplify ==> 0 ( Could apply "simplify" in this case ? )
Hello Everyone.
1i*ln(1/[e^pi]^1i) simplify ==> -pi and 1i*ln(1/[e^pi]^1i) = -pi simplify ==> 0 ( Could apply "simplify" in this case ? )
Thanks in advance.
Regards.