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:

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

Thus

\(\ds \cmod {1 - \frac {\map \ln {1 + z} } z}\)

\(=\)

\(\ds \cmod {\sum_{n \mathop = 2}^\infty \frac {\paren {-1}^n z^{n - 1} }n}\)

Linear Combination of Convergent Series

\(\ds \)

\(\le\)

\(\ds \sum_{n \mathop = 2}^\infty \cmod {\frac {\paren {-z}^{n - 1} } n}\)

Triangle Inequality for Series

\(\ds \)

\(\le\)

\(\ds \sum_{n \mathop = 2}^\infty \frac {\cmod z^{n - 1} } 2\)

$n \ge 2$

\(\ds \)

\(=\)

\(\ds \frac {\cmod z / 2} {1 - \cmod z}\)

Sum of Infinite Geometric Sequence

\(\ds \)

\(\le\)

\(\ds \frac 1 2\)

$\cmod z \le \dfrac 1 2$

By the Triangle Inequality:

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

$\blacksquare$

Also see

• Bounds of Natural Logarithm
• Bounds for Complex Exponential

Sources

• 1973: John B. Conway: Functions of One Complex Variable ... (previous) ... (next) $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S 5$: Weierstrass Factorization Theorem: $5.3$