come on, come on, what ya' waitin' for? Top one is the same except the argument change is the negative of the bottom so we got zero accumulated argument change over top and bottom or is that -pi thing up there not too easy to see? Look at those later. Anyway, gotta' find \Delta \arg=2\pi somewhere else (cus' there's a zero in there, check the theorem). How about the left contour? Got z=-m+iy so we end up with:
f(-m+iy)=\left(e^{-m} \cos(y)+m\right)+i\left(e^{-m}\sin(y)-y\right)
When m is very large, say in the limit as m goes to infinity, the real part tends to a very large number (m), and the imaginary component goes from 0 to -2\pi i. But in the limit as m goes to infinity, that change in argument is an infinistessimal so the change in argument there is zero.
You know the zeta function is really hard to do this to. Anyway, must be the last contour. Can we find, in the limit as m goes to infinity, an accumulated change in argument of 2pi since what, we got one zero in there?
What if we analyze:
f(m+iy)=\left(e^{m} \cos(y)-m\right)+i\left(e^{m}\sin(y)-y\right)
for very large m and y as it goes from 0 to 2pi (k=0)? How could you prove the change in argument of f(m+iy) over that contour as m goes to infinity tends to 2\pi?
Edit: When I say the top, bottom, left and right contour I mean the box contour with top, bottom, left, and right legs with left bottom corner at (-m,0) and top right corner at (m,2pi).
