$|\log (1 + z)| \leq 2 |z|$ Complex inequality

Question

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!

Answer 1

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 $$ |\log(1+z)|=\left|\sum^{\infty}_{n=1}\frac{(-1)^nz^n}{n}\right|\leq|z|\sum^{\infty}_{n=0}\frac{|z|^n}{n+1}<|z|\left(\frac{1}{1}+\frac{1}{2\cdot 2}+\frac{1}{3\cdot 2^2}+\frac{1}{4\cdot 2^3}+\cdots\right)=|z|C, $$ 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