How do you prove the following inequality concerning complex Logarithms? If $0<|w|<1/2$, then $2|w|/3<|\operatorname{Log}(1+w)|$ using power series and modulus inequalities.
 A: Though it is probably not the cleanest method, it is possible to use the minimum modulus principle and explicitly calculate the minimum value of $|\log(1+w)/w|$ on $|w| = 1/2$.
As noted by AD., the function
$$
f(w) = \log(1+w)/w
$$
is analytic on $|z| < 1$.  Further, it is nonconstant and zero-free there, so we can appeal to the minimum modulus principle to conclude that
$$
|f(w)| > \min_{|z| = 1/2} |f(z)|
$$
for $|w| < 1/2$.  Now,
$$
\left|\log\!\left(1+\frac{e^{i\theta}}{2}\right)\right|^2 = \frac{1}{4}\left[\log\!\left(\frac{5}{4}+\cos\theta\right)\right]^2 + \arctan\!\left(\frac{\sin\theta}{2+\cos\theta}\right)^2
$$
is symmetric about $\theta = \pi$.  By beating the derivatives senseless and using the fact that $x \operatorname{arccot} x < 1$, it is possible to show that this function is strictly increasing on $(0,\pi)$.  Thus
$$
\min_{|z| = 1/2} |f(z)| = 2 \min_{0 \leq \theta \leq \pi/2} \left|\log\!\left(1+\frac{e^{i\theta}}{2}\right)\right| = 2\log\!\left(\frac{3}{2}\right) > \frac{2}{3},
$$
from which the result follows.
