This is the full question.
Show that $|\int_{|z|=R}\frac{\log(z)}{z^2}dz| \leq 2\pi(\frac{\pi + ln(R)}{R})$ and then use L'Hopital's rule to show that the value of this integral tends to $0$ as $R$ tends to infinity.
This is what I have so far:
$\gamma(t) = Re^{it}$
$\gamma'(t) = Rie^{it}$
$|\gamma'(t)| = R(1)(1)$
$length(\gamma) = \int_{-\pi}^{\pi}|\gamma'(t)||dz| = 2\pi R$$$|\int_{|z|=R}\frac{\log(z)}{z^2}dz| \leq \int_{|z|=R}|\frac{\log(z)}{z^2}||dz|$$
$$ ................\leq \int_{|z|=R}\frac{|\log(z)|}{|z^2|}|dz| $$ $$.........................\leq \int_{|z|=R}\frac{|\ln|z| + Arg(z)|}{R^2}|dz| $$ $$...................\leq 2\pi R \frac{|ln|z| + Arg(z)|}{R^2} $$ $$.............\leq 2 \pi \frac{\ln(R) + \pi}{R}$$
I am confused about going from $|ln|z| + Arg(z)|$ to $|ln(R) + \pi|$. I assume that since we are dealing with the principle argument, that we are going from $-\pi$ to $\pi$, and since we are going counter clockwise, this just means $\pi$. But, I'm not too sure about the natural log part as well as what happens to modulus symbol (|...|).
I wanted to also see if the following way of thinking was correct. The most that $|z|$ can be is $R$, which means $|z|^2 = R^2$, and so $\frac{1}{|z|^2} = \frac{1}{R^2}$.
$|ln|z| + Arg(z)|\leq |\ln |z||+|Arg(z)|=|\ln R|+|Arg (z)|=\ln R+|Arg (z) | \leq ln(R) + \pi $ assuming that $R>1$ so $\ln R$ is positive.
For the last part of the question we are integrating over $|z|=R$ so $|z|$ is actually fixed. Ne need to maximize it.