Sorry for the bad notation. Also had to steal symbols like 𝜋 from other places...
Okay, there are two ways of looking at this question. I'll start with the more likely one (that it's not a closed contour and that you don't need Cauchy's integral theorem) first.
Here, we have take the path integral from 5 to 3i, treating the complex plane like multivariable calc.
∫1/z dz= ln |z|=Log r+i*theta.
The second part follows from forming a single-valued branch cut of the logarithm function (for one to one mapping instead of having a many-to-one function) by removing a ray (by convention the negative real axis).
The many-to-one thing can clearly be seen in that exp(z + 2π i n) = exp(z) · E2 Pi*i*n = exp(z)
I suppose the branch cut kind of uses Cauchy's integral formula by effectively removing the branch point at z=0... since it effectively creates a contour that "cuts" the branch point and a ray out by deforming a curve around these points.
Thus, our line integral is simply:
Ln |3i|-Ln |5|=(Log[3]+i*Pi/2)-(Log[5]+i*0)=Log[3/5]+i*𝜋/2
I strongly believe you're referring to Cauchy's integral theorem in this case, and I ignored the part about limits since I thought you were learning about CIT, and this was some type of trick question. Assuming this, we obtain relatively easily from Cauchy's Integral Theorem: http://mathworld.wolfram.com/CauchyIntegralTheorem.html,
which basically tells us that along any closed contour that has an inside which is completely differentiable (analytic), the integral will always evaluate to zero. This means that if we had a pole (singularity), we can expand the outward boundary so long as we don't pass through any other poles all the while maintaining the same integral around our new contour. (which can have a non-zero value if your boundary contains a pole).
Then, the math becomes very simple.
If we do not go around our pole at 0, the contour integral is 0 by CIT.
On the other hand, if we do:
By setting z=R*Ei*theta then dz=i*R*Ei*theta dtheta
we obtain:
∫C1/z dz where C is some contour containing the pole (in our case z=0)
= ∫C( i*R*Ei*theta)/R*Ei*theta dtheta
By CIT, this contour will be equivalent to the unit circle around our pole so:
i*Integral[dtheta, {theta,0,2𝜋}]= (i*2𝜋)-(i*0)=2 𝜋*i
