The Integral of 1/z

Complex Analysis · Axiom Academy

One contour integral, equal to , is the seed of all of residue theory — here is why it refuses to be zero. 1. Once Around the Unit Circle Take the unit circle, traversed counterclockwise, and parametrize it by z(t) = e^ it . Then z'(t) = ie^ it , and since we get f(z(t)) = e^ -it . Multiply and the exponentials cancel exactly — the integrand collapses to the constant i . parametrize, differentiate, substitute integrating the constant i over a -long trip 2. The Singularity Is What Counts For tame functions like z , z^2 , or e^z , every closed-loop integral is 0 . So why is different? Because fails to be analytic at z=0 , and that bad point sits inside the unit circle. Cauchy's theorem only zeros out a loop when the function is analytic everywhere the loop encloses — and here it is not. 0 lies inside |z|=1 , so Cauchy's theorem does not apply and the integral is . Around |z-2|=1 the point 0 is outside, is analytic within, and the integral is 0 . With z(t)=Re^ it the radius R cancels — the value is for every circle about the origin. counts how many times wraps 0 ; one CCW lap gives 1 . It is tempting to say , so the loop integral should vanish. The catch: is multi-valued . Going once around the origin, does not return to its start — it jumps by , which is precisely the value we computed. 3. Among All Powers, Only z^ -1 Survives

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