# Bounds for Complex Logarithm

## Theorem

Let $\ln$ denote the complex logarithm.

Let $z \in \C$ with $\cmod z \le \dfrac 1 2$.

Then:

$\dfrac 1 2 \cmod z \le \cmod {\map \ln {1 + z} } \le \dfrac 3 2 \cmod z$

## Proof

By definition of complex logarithm:

$-\map \ln {1 + z} = \displaystyle \sum_{n \mathop = 1}^\infty \frac {\paren {-z}^n} n$

Thus

 $\displaystyle \cmod {1 - \frac {\map \ln {1 + z} } z}$ $=$ $\displaystyle \cmod {\sum_{n \mathop = 2}^\infty \frac {\paren {-1}^n z^{n - 1} }n}$ Linear Combination of Convergent Series $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = 2}^\infty \cmod {\frac{\paren {-z}^{n - 1} } n}$ Triangle Inequality for Series $\displaystyle$ $\le$ $\displaystyle \sum_{n \mathop = 2}^\infty \frac {\cmod z^{n - 1} } 2$ $n \ge 2$ $\displaystyle$ $=$ $\displaystyle \frac {\cmod z / 2} {1 - \cmod z}$ Sum of Geometric Sequence $\displaystyle$ $\le$ $\displaystyle \frac 1 2$ $\cmod z \le \dfrac 1 2$

By the Triangle Inequality:

$\displaystyle \frac 1 2 \le \cmod {\frac {\map \ln {1 + z} } z} \le \dfrac 3 2$

$\blacksquare$