Evaluating a Contour Integral of log(z) | Explaining Branch Cuts

[eyesontheball1]

Is it possible to evaluate the integral of log(z) taken over any simple closed contour encircling the origin? I don't fully understand how singularities on branch cuts should be treated when integrating over contours encircling such singularities. Are residues applied? Can someone explain this to me? Thanks!

 

[HallsofIvy]

Looking at your specific example, ln(z), by convention we take the branch cut along the negative real axis or $\theta= \pi$ when [itex]z= re^{i\theta}[/tex]. And, of course, $$ln(z)= ln(r)+ i\theta$$. So integrating across a branch cut results in adding [tex]2\pi n i$$for <b>some</b> integer n. That is the difficulty with integrating over branch cuts- the integral value jumps by some multiple of a constant.$$[/tex][/itex]

 

[Svein]

If you put z = r*(cos(φ) + i*sin(φ)), log(z) = ln(r) + i*φ (since both sine and cosine are periodic with period 2π, there are several values of φ we can use in the expression).

Thus ∫log(z) for |z|=R, is equivalent to ∫(ln(R) +i*φ)dφ where the integration limits are 0 and 2π. The rest is left as an exercise for the student...