# I Domain of single-valued logarithm of complex number z

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1. Nov 28, 2016

### goodphy

Hello.

Let's have any non-zero complex number z = re (r > 0) and natural log ln applies to z.

ln(z) = ln(r) + iθ. In fact, there is an infinite number of values of θ satistying z = re such as θ = Θ + 2πn where n is any integer and Θ is the value of θ satisfying z = re in a domain of -π < Θ < π. So, ln(z) can be written as ln(z) = ln(r) + i(Θ + 2πn).

ln(z)
is multiple-valued function as each Θ + 2πn with different n results in different ln(z), although Θ + 2πn are essentially same angle between radial line from the origin to z and the positive real axis in the complex plane (in other word, Θ + 2πn indicates same z).

In order to make single-valued and continuous function, the domain of θ needs to be restricted somehow. The textbook of complex analysis says a way to restrict domain of the multiple-valued function ln(z) to make the single-valued function F is lik this; F = ln(z) = ln(r) + iθ (α < θ < α + 2π).

My question is why the domain is restricted by (α < θ < α + 2π), instead of (α ≤ θ < α + 2π)? I think later domain also can be used to F. F is continous from θ = α itself to the point infinitesimally close to α + 2π and is single-valued over this domain.

I think tihs is important question for me as I'm trying to understand the concept of the branch cut and branch point for the integral on a contour in the complex plane.

2. Nov 28, 2016

### FactChecker

You are correct. You can include one of the end points of the range of θ

3. Dec 4, 2016

### goodphy

I see. I was worrying whether the textbook was misprinted. Other mathematical physics book says the domain of the single-valued function is F(z) = ln(z) = ln(r) + iθ is α ≤ θ < α + 2π.

I think my book was misprinted:)