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.