Bounds for Complex Logarithm

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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 Progression
\(\displaystyle \) \(\le\) \(\displaystyle \frac 1 2\) $\cmod z \le \dfrac 1 2$

By the Triangle Inequality:

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

$\blacksquare$


Also see


Sources