Analyticity of inverse tan

[Tangent87]

The function Tan^-1z is defined by:

$$Tan^{-1}z=\int_0^z \frac{dt}{1+t^2}$$

where the path of integration is a straight line. For what region of the complex plane is Tan^-1z analytic?

I'm not too sure about this. I can see that the integrand is analytic everywhere except at t=+i and t=-i so does that mean that Tan^-1z is analytic as long as we don't take a path of integration through those points? Or is Tan^-1z analytic everywhere except at z=+i and z=-i?

 

[mighty2000]

I can help clarify this question for you. First, let's define what it means for a function to be analytic. A function is analytic at a point if it has a well-defined derivative at that point. In other words, it is smooth and continuous at that point.

Now, in the case of Tan-1z, we can see that the integrand is indeed analytic everywhere except at t=+i and t=-i. This is because at these points, the denominator becomes zero and the function is not well-defined.

However, this does not necessarily mean that Tan-1z is not analytic at those points. It just means that the path of integration cannot pass through those points. In other words, Tan-1z is analytic for all points in the complex plane except for the points z=+i and z=-i.

To summarize, Tan-1z is analytic for all points in the complex plane except for z=+i and z=-i. This is because the integrand is analytic everywhere except at t=+i and t=-i, and the function is not well-defined at those points. So, as long as the path of integration avoids these points, Tan-1z will be analytic.