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.