Why is this way of computing imaginary arctangents wrong?

  • #1
181
13
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?
 

Answers and Replies

  • #2
34,665
6,379
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.
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)
David Carroll said:
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))
arctan(x) is a constant, so its derivative is 0.
David Carroll said:
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?
 
  • #3
181
13
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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