Bound in Complex Analysis Can someone direct me towards the right way to approach this problem? Show $$\displaystyle \left|\int_{|z|=R} \frac{Log{z}}{z^2} dz\right| \leq 2\sqrt{2}{\pi}\frac{\log{R}}{R},\; \text{ for } R>e^{\pi}.$$
Edit: I'm considering the ML estimate, so L would be $2\pi R$ and I'm still debating M. The max for the $\frac{1}{z^2}$ is using R, since it's the smallest value... But I'm still quite lost.
 A: Use that
$$
|\operatorname{Log}z|=|\log|z|+i\operatorname{arg}z|\le\sqrt{(\log|z|)^2+\pi^2}.
$$
I am assuming that $-\pi<\operatorname{arg}z\le\pi$.
A: Note that
$
\begin{align}
  \biggr|\frac{\log z}{z^{2}}\biggr| = \biggr|\frac{\log|z| + i\operatorname{Arg}(z)}{z^{2}}\biggr| &\leqslant\frac{\bigr|\log |z| + i\pi\bigr|}{|z^{2}|} \\
  &= \frac{|\log R + i\pi|}{R^{2}} \\
  &\leqslant \frac{\sqrt{2\log^{2} R}}{R^{2}} \\
  &= \sqrt{2}\frac{\log R}{R^{2}}
\end{align}
$
Thus we have
$
\begin{align}
  \biggr|\oint_{|z| = R}\frac{\log z}{z^{2}}\,\mathrm{d}z\biggr| &\leqslant\sqrt{2}\frac{\log R}{R^{2}}\cdot 2\pi R\:\:\text{ ( by ML estimate ) } \\
  &= 2\sqrt{2}\pi\frac{\log R}{R}
\end{align}
$
A: Use that
$${{\rm Log}\,z\over z^2}=-{d\over dz}{{\rm Log}\,z+1\over z}\qquad\bigl(|{\rm Arg}\,z|<\pi\bigr)\ .$$
Cutting $|z|=R$ up at the point $-R\in{\mathbb C}$ and passing to the limit this allows to represent the exact value of the integral as a difference:
$$\int_{\partial D_R}{{\rm Log}\,z\over z^2}\>dz=-{{\rm Log}\,z+1\over z}\biggr|_{Re^{-i\pi}}^{Re^{i\pi}}={(\log R-i\pi)+1\over -R}-{(\log R+i\pi)+1\over -R}={2i\pi\over R}\ .$$
It follows that
$$\left|\int_{\partial D_R}{{\rm Log}\,z\over z^2}\>dz\right|={2\pi\over R}\ ,$$
which certainly beats your estimate.
