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We know that $\log z = \log\vert z\vert + i\arg(z)$. $\log z$ and $\log\vert z\vert$ are only different by a bounded function. (More specifically by $i\pi$ since $\vert i\arg(z)\vert\le\pi.)$

Now however I'm interested in the case $$\log(1+c\vert z\vert ^2)\quad\text{and}\quad \vert \log(1+c z ^2)\vert,$$ where $c$ is a positive real number and $z$ complex, more spefically in the right half plane. Now I wonder if these two also only differ by a constant (or a bounded funcction) and if this is the case, how I can prove it.

bochner.martinelli
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1 Answers1

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The answer is no. Consider the difference between the two:

$$\begin{eqnarray} \log(1 + cz^2) - \log(1 + c|z|^2) & = & \log|1 + cz^2| + i \arg(1 + cz^2) - \log(1 + c|z|^2) \\ & = & i \arg(1 + cz^2) + \log \left(\frac{|1 + cz^2|}{1 + c|z|^2} \right) \end{eqnarray}$$

If we then look at the fraction within the logarithm, the denominator is always going to be $\geq 1$, but as $z^2 \rightarrow \frac{1}{c}$, the numerator is going to approach zero, meaning that we can make the whole thing arbitrarily small, and so when we take the logarithm it will tend towards negative infinity.

ConMan
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