TITLE: $|\log (1 + z)| \leq 2 |z|$ Complex inequality
QUESTION [5 upvotes]: Prove that for $|z|\leq0.5$, $|\log (1 + z)| \leq 2 |z|$. 
I know that $|\log (1 + z)|=|\log|1+z|+i\arg(1+z)|$ and $|\arg(1+z)|\leq\pi/6$ for $|z|\leq0.5$, but then I don't know how to proceed. It seems that the it attains "=" when $z=0$? Thanks!

REPLY [4 votes]: For complex $z$, with $|z|\leq1$ and $z\neq 1$, we have 
$$
\log(1+z)=\sum^{\infty}_{n=1}\frac{(-1)^{n-1}}{n}z^n.
$$
Hence for $|z|\leq\frac{1}{2}$ we get
$$
|\log(1+z)|=\left|\sum^{\infty}_{n=1}\frac{(-1)^nz^n}{n}\right|\leq|z|\sum^{\infty}_{n=0}\frac{|z|^n}{n+1}<|z|\left(\frac{1}{1}+\frac{1}{2\cdot 2}+\frac{1}{3\cdot 2^2}+\frac{1}{4\cdot 2^3+\ldots}\right)=|z|C,
$$
where 
$$
C=\sum^{\infty}_{n=0}\frac{1}{(n+1)2^n}<\sum^{\infty}_{n=0}\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=2.
$$
QED