# Why is this way of computing imaginary arctangents wrong?

• David Carroll
In summary, the conversation discusses a method for computing the arctangent with imaginary arguments. It presents a conditional equation that is not an identity, and goes on to show the steps for solving it. However, there is an error in the process, as arctan(x) is a constant and its derivative is 0, not -y. The conversation ends with the realization of this error.

#### David Carroll

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?

David Carroll said:
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.
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 ) in the above?

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.

## 1. Why do we need to compute imaginary arctangents?

Imaginary arctangents can be used to solve problems involving complex numbers, such as finding the angle of a complex number in the complex plane.

## 2. What is the correct way to compute imaginary arctangents?

The correct way to compute imaginary arctangents is by using the inverse tangent function for complex numbers, also known as the "atan2" function.

## 3. Why is this way of computing imaginary arctangents wrong?

This way of computing imaginary arctangents is wrong because it assumes that the imaginary part of the complex number is equal to the radius of the unit circle, which is not the case for all complex numbers.

## 4. What are the consequences of using the wrong method to compute imaginary arctangents?

Using the wrong method to compute imaginary arctangents can lead to incorrect results and can cause errors in further calculations involving complex numbers.

## 5. Can this wrong method be used in any situation?

No, this wrong method should not be used in any situation as it can lead to incorrect results and errors in calculations involving complex numbers.