- #1

goodphy

- 216

- 8

Let's have any non-zero complex number

*z = re*(

^{iθ}*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

^{iθ}*where*

*θ = Θ + 2πn**n*is any integer and

*is the value of*

*Θ**θ*satisfying

*z = re*in a domain of -π <

^{iθ}*. So,*

*Θ*< π*ln(z)*can be written as

*ln(z) = ln(r) + i(*

ln(z)is multiple-valued function as each

*).**Θ + 2πn*ln(z)

*with different n results in different*

*Θ + 2πn**ln(z)*, although

*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*

*Θ + 2πn**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 continuous 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.