Bounds for Complex Logarithm

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Theorem

Let $\log$ denote the complex logarithm.

Let $z\in\C$ with $|z|\leq\frac12$.


Then $\frac12|z| \leq |\log(1+z)| \leq \frac32|z|$.


Proof

By definition of complex logarithm:

$-\log(1+z) = \displaystyle \sum_{n\mathop=1}^\infty\frac{(-z)^n}n$

Thus

\(\displaystyle \left\vert 1-\frac{\log(1+z)}z \right\vert\) \(=\) \(\displaystyle \left\vert \sum_{n\mathop=2}^\infty \frac{(-1)^{n}z^{n-1} }n \right\vert\) Linear Combination of Convergent Series
\(\displaystyle \) \(\leq\) \(\displaystyle \sum_{n\mathop=2}^\infty\left\vert \frac{(-z)^{n-1} }n \right\vert\) Triangle Inequality for Series
\(\displaystyle \) \(\leq\) \(\displaystyle \sum_{n\mathop=2}^\infty \frac{\vert z\vert^{n-1} }2\) $n\geq2$
\(\displaystyle \) \(=\) \(\displaystyle \frac{\vert z\vert /2}{1- \vert z\vert}\) Sum of Geometric Progression
\(\displaystyle \) \(\leq\) \(\displaystyle \frac12\) $\vert z\vert \leq \frac12$

By the Triangle Inequality:

$\displaystyle \frac12 \leq \left\vert \frac{\log(1+z)}z \right\vert \leq \frac32$

$\blacksquare$


Also see


Sources