# 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.

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It doesn't appear true for $w=0$ –  Ross Millikan Sep 12 '11 at 17:48
What's the power series for $\operatorname{Log}(1+w)$? (That's a hint, not a question:) –  marty cohen Sep 12 '11 at 19:45

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.

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I am willing to include the proof that the function is strictly increasing if anyone would like. I left it out only because it's tedious. –  Antonio Vargas May 16 '12 at 7:22
Yes that is the way, removed my hints since there was an error. (+1) –  AD. May 16 '12 at 19:51