I was doing some things in my head the other day (these moments usually don't come out so well ). And I "came up" with the following way to compute the arctangent with imaginary arguments. Consider the identity x2 = - 2, where x = ± i√2. Now rearrange this identity to x2 + 1 = -1 Invert the equation to get the reciprocals 1/(x2 + 1) = - 1 Multiply both sides by arctan(x) arctan(x)(1/(x2 + 1)) = - arctan(x) Let y = arctan(x) Then we have the differential equation yy' = - y (Because 1/(x^2 + 1) is the derivative of arctan(x)) Divide both sides by y to get y' = - 1 Now integrate both sides ∫dy = - ∫dx The result is y = - x But since y = arctan(x), it follows that arctan(x) = - x And since we know that x = ± i√2, it follows that arctan(i√2) = - i√2 and arctan(-i√2) = i√2 I know I must have done something wrong here. Could someone please point out the errors (which I'm sure are many ) in the above?