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Why is this way of computing imaginary arctangents wrong?

  1. Dec 10, 2014 #1
    I was doing some things in my head the other day (these moments usually don't come out so well :rolleyes:). 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 o0)) in the above?
     
  2. jcsd
  3. Dec 10, 2014 #2

    Mark44

    Staff: Mentor

    The above is not an identity. It is a conditional equation that is true only for two values of x. Here are two examples of identities:
    1. x2 + 2x + 1 = (x + 1)2
    2. sin(2x) = 2sin(x)cos(x)
    arctan(x) is a constant, so its derivative is 0.
     
  4. Dec 10, 2014 #3
    Thanks, Mark44. I actually just realized the error about 17 seconds after I clicked on "Create Thread".

    I definitely should not quit my day welfare.
     
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