Holomorphic 1/z

Hmm... last last time when I saw this:

Complex analysis is not real analysis. Complex is actually fun, its just another class where you manipulate numbers and variables and see cool things.

I thought you had actually studied some of these types of problems. (Not trying to offend or anything.)

 Feel free to skip this

Yesterday, as I was eating dinner, I realized that my TA walked off with my vector calculus textbook after discussion, thus rendering me unable to complete my homework. To remedy this, I tried to search online for my textbook (I ended up downloading a Spanish version of the textbook and typed it into Google translate to work the problems. :|) In the process of searching, I stumbled upon Marsden and Hoffman's Basic Complex Analysis 3rd Edition.

Previously, I had Stein and Shakarchi on my desktop, but I didn't really look through it, since the topics it discusses weren't in my range of knowledge (or range of interest, for some of them). I haven't studied Fourier's transform, Gamma/Zeta/Theta functions, prime numbers, conformational mappings, elliptic functions, etc. in full detail, and that's pretty much all of Stein and Shakarchi. I only kept the book on my desktop and glimpsed through it when I'm bored, because I didn't really have another book from which to view this subject (and we all know that Wikipedia is crap for math).

Anyways, I've known for a while now that ∮1/z dz = 2πi, even though I never really understood how since I didn't know how to parametrize in complex variables. Yesterday, I briefly looked through the first two chapters of Marsden and Hoffman and asked myself the simple question: how do I show that the deformation theorem holds? By showing that different paths give the same value.

And so I wanted to calculate ∮1/z dz over three different paths, and the latter two unfortunately failed. Maybe I'm just parametrizing wrong, or, like hadriel pointed out, I need to somehow get rid of the log function to extract the actual value.

Today, I realized that I could probably simplify this by factoring out e^(iθ. However, as I show below, this causes some strange behavior (here, log is of base e):

g(θ = log (3/2 e^(iθ + 1/2 e^(-iθ)
= log(e^(iθ (1/2) (3 +e^(-2iθ), [0,2π]
= log(e^(iθ) - log(2) +log(3+e^(-2iθ), [0,2π]
= iθ [0,2π]
= 2πi

While that may seem correct now, notice that if I factor out e^(-iθ instead, I obtain:
g(θ = log (3/2 e^(iθ + 1/2 e^(-iθ)
= log(e^(-iθ (1/2) (3e^(2iθ + 1), [0,2π]
= log(e^(-iθ) - log(2) + log(3e^(2iθ + 1), [0,2π]
= -iθ [0,2π] (if log(3e^(2iθ + 1), [0,2π] = 0)
= -2πi

Actually, I can factor out exponential anything. For example, by factoring out e^(2iθ within the log, I erroneously obtain 4πi instead of 2πi. If factoring them out "any way" gives correct solutions, then the above factorization (bolded) must show that

log(3e^(2iθ + 1), [0,2π] = 4πi

which doesn't really make sense.

Also, putting the following into Mathematica:

log(3/2*e^(i*2pi)+1/2*e^(-i*2pi))-log(3/2*e^(i*0)+1/2*e^(-i*0))

returns 0, when it should return the "correct" solution of 2πi. In fact, putting in the simple identity log(e^(i*2pi))=2πi causes Mathematica to erroneously obtain 0. Also, another problem is that even if my first simplification is correct, I'm unable to guarantee that:

log(3+e^(-2iθ), [0,2π] = 0

I'm still completely lost with the third one. There doesn't seem to be a way to kill the log without an exponential term. Hopefully, Noah knows how to cure this mess.