First of all, you haven't thought about what curve C' looks like, but it is actually like this.
C' is not a simple closed curve. When z moves counterclockwise along |z| = 2, w = e^(2z) bypasses-1 moves counterclockwise along c' for two weeks.
So ∮ {c'}1(1+w) dw = 2 ∮ {| w | = e 4}1/(1+w) dw.
In fact, directly using residue theorem here, we can calculate:
∮{c'} 1/( 1+w)dw = 2 ^ 2πI RES( 1/( 1+w),- 1) = 4πi
Instead of residue theorem, we try to turn it into a real integral.
It's not impossible, but you didn't notice the multivalued nature of Ln(z).
According to your idea, you will get the wrong conclusion that ∮ {| z | =1}1/zdz = ln (e (2 π i))-ln (e 0) = 0.
The correct result is ∮{|z| = 1} 1/z dz =? ∫{0,2π} e^(-iθ) d(e^(iθ)) = i ∫{0,2π} dθ =? 2πi。
The reason for the error is that for the same z, Ln(z) may differ by an integer multiple of 2 π i. 。
Finally, two suggestions are given:
First of all, don't change Yuan easily for this topic.
For this problem, |z| = 2 is a familiar circle, but it is not easy to figure out what C' looks like.
In fact, just directly find the pole and residue of e z/cosh (z) in |z| = 2.
It is not difficult to find two poles z =? π I/2, the remainder is 1.
The second is that complex integrals should not always be converted into real integrals.
Generally, it can't be calculated, and the above-mentioned multi-value problems may occur during calculation.
It is not a pity to have such a useful tool as residue theorem.