Holomorphic 1/z

How do I select a branch cut? I can't even find anything online to visualize complex functions.

got internet back!!!!!!!!

 Spoiler

for example, i can choose

t e [0,8]
z = -1+i(1-t) [0,2]
z = (t-3)-i [2,4]
z = 1+i(t-5) [4,6]
z = (7-t)+i [6,8]

then

∮1/z dz = ∫ -i/(-1+i(1-t)) dt [0,2] + ∫ 1/((t-3)-i) dt [2,4] + ∫ i/(1+i(t-5)) dt [4,6] + ∫ -1/((7-t)+i) dt [6,8]
= {ln(-1-i) - ln(-1+i)} + {ln(1-i) - ln(-1-i)} + {ln(1+i) - ln(1-i)} + {ln(-1+i) - ln(1+i)}
= iπ{5/4 - 3/4} + iπ{7/4-5/4} + iπ{1/4-7/4} + iπ{3/4-1/4} = 0

now then... it's 0. what? well yes. why? branch cut. we crossed it. that's gonna bugger you. up to now, i haven't even declared where my branch cut is. branch cut makes things single valued, and you can't cross it with your integral's path. hence, in method 1, you might see it in a textbook as [0,2π). usually you make a branch cut as convenient as possible. physics people might use [-π,π sometimes, math people kinda don't care as long as you mention it. taking the interval [0,2π implies taking the branch cut along the positive real axis. with our square geometry, it's convenient to cut from 0 to a corner of the square. pick any one. i'll choose the -1,i corner, since that's where t starts. cutting there allows us to define a domain for e^iθ, in this case it's [-π5/4, π3/4). so now take a look at that last line again:

= iπ{5/4 - 3/4} + iπ{7/4-5/4} + iπ{1/4-7/4} + iπ{3/4-1/4}

now that we have taken our branch cut, anything outside of the domain can't be right. so iπ7/4 -> -iπ1/4 and iπ5/4 -> -iπ3/4. (these won't actually matter, but it's good practice to remember they don't really exist with this branch cut). so if changing those doesn't matter, what are we forgetting? we're forgetting we can't cross the branch cut. we START just "below" the branch cut and end just "above" it. that means our starting value of iπ3/4 is wrong and we need to swap it for the clockwise moving way of labeling it: -iπ5/4.

= iπ{-3/4 + 5/4} + iπ{-1/4 + 3/4} + iπ{1/4+1/4} + iπ{3/4-1/4}
=i2π

one more thing... like i said, iπ3/4 is outside of our domain, why did i still use it once? use the fundamental theorem of calculus to evaluate the last integral at its endpoints, since it's just a riemann integral we can do that, but iπ3/4 is outside of our domain, so call that endpoint "a" instead and take the limit as a->iπ3/4 and it converges nicely and all that (just to be sure i don't try to slip one past you, it's the typical justification)

hope this helps

selecting a branch cut is up to you. you'll need one any time you have singularities in the complex plane. branch cuts represent lines you don't cross, ever, when searching for a path to integrate over. they keep everything single valued. and WHAT they look like doesn't matter. for example, in ^ i could have chosen my branch cut to spiral around the origin 20 times, gone through (-1,i), and then done a saw function in any direction out to infinity (long as it doesn't cross my path again). kind of pathological, but it makes more sense when say... you look at normal modes of acoustic waves and you have some strangely placed singularities you need to evaluate the strength of with something like residue theorem.