I have the integral,
$$I(R) = \int_{C_R}\frac{1}{z(z-1)^2} dz$$
with the property that
$$\left|\frac{1}{z(z-1)^2}\right| \leq \frac{1}{R(R-1)^2} \quad |z|=R>1$$
Where $C_r$ is the contour defined by the circle of radius $R>1$ so $C_r $ can be parametrized as $z(\theta)=Re^{i\theta}$
The goal is to show that $I(R)$ is constant for $R>1$.
I can split $C_R$ into two contours ($0\notin L_1, 1\in L_1$, $L_2=C_R \setminus L_1$) and then use Cauchy's Integral Formula and immediatly show that $I(R)=0$:
$$I(R) = \int_{L_1}\frac{1/z}{(z-1)^2} dz + \int_{L_1}\frac{1/(z-1)^2}{z} dz= 0 $$
But apparently there's some way of showing that $I(R)$ is constant without computing it?!
I don't see how. Clearly the function inside the integral is not entire and if I split it into two contours, the functions in the numerator are analytic within their respective contours but their maximums are achieved outside those contours so it does not follow that they are constant either.
The problem is that the point at which they are maximum is the point at which they stop being analytic.
So how can one imply that $I(R)$ is constant without showing that $I(R)=0$?
I guess you could split $C_R$ into $C_1$ and the contour enclosed by $C_R$ and $C_1$ (let's call it $L(R)$) where there is a line going from one circle to the other somewhere on the circumference. This makes the domain enclosed by the second contour, simply connected.
$$I(R) = \int_{C_1}\frac{1/z}{(z-1)^2} dz + \int_{L}\frac{1/(z-1)^2}{z} dz\quad where \quad \int_{L(R)}\frac{1/(z-1)^2}{z}dz=0$$
Therefore the integral over the second contour will be $0$ since the function inside the integral is analytic everywhere outside the circle of radius $1$. Then the integration would be constant. Is this reasoning valid? Also I'm not sure about the whether I should add or subtract the integrals. If I define both contours as positively oriented then I should subtract them, right?
The integral of a rational function like that around a closed contour will always be written as a sum of residues, so it only depends on the poles inside the contour. Therefore, once the contour encloses all the poles, the function does not depend on $R$ anymore. You don't really have to calculate it, you just have to make sure that $R$ is big enough so that the contour has enclosed all the poles