In a real variable integral, it is true that $\oint f(x)dx\leq \oint g(x)dx$ if $|f(x)|\leq |g(x)|$ for all x. But i think the similar proposition doesn't hold for the case of complex plane contour integral , $|\oint f(z)dz|\leq |\oint g(z)dz|$ if $|f(z)|\leq |g(z)|$ for all z .
My qusetion:
Is there any possible ways or conditions to compare two contour integral's absolute value? It doesn't have to be as clear as above inequality. I'm just wondering whether such a similar inequality exists.
Thank you for reading my qusetion and have a nice day!
If $f$ and $g$ are analytic on $\mathbb C$ except for isolated singularities, and $|f(z)| \le |g(z)|$ whenever both are defined, then $\left|\int_C f(z)\; dz\right| \le \left|\int_C g(z)\; dz\right|$ for any path $C$ for which the integrals are defined.
The catch is that this is trivial. The function $h(z) = f(z)/g(z)$ has only isolated singularities which are all removable, and after removing them you have a bounded analytic function, which must be a constant. So $f(z) = c g(z)$ for some constant with $|c|\le 1$, and then $\int_C f(z)\; dz = c \int_C g(z)\; dz$.