Logarithmic Derivative of Product of Analytic Functions
Jump to navigation
Jump to search
This article, or a section of it, needs explaining, namely:
Link to a clarifying definition of what $\left({f g}\right)$ is -- presumably it's the pointwise product, but in the context there's nothing to say it can't be $\left({f \circ g}\right)$, that is, $f$ of $g$ of $z$.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
Theorem
Let $D \subset \C$ be open.
Let $f, g: D \to \C$ be analytic.
Let $z \in D$ with $f \left({z}\right) \ne 0 \ne g \left({z}\right)$.
Then:
- $\dfrac{\left({f g}\right)' \left({z}\right)} {\left({f g}\right) \left({z}\right)} = \dfrac{f' \left({z}\right)} {f \left({z}\right)} + \dfrac {g' \left({z}\right)} {g \left({z}\right)}$
Link to a clarifying definition of what $\left({f g}\right)$ is -- presumably it's the pointwise product, but in the context there's nothing to say it can't be $\left({f \circ g}\right)$, that is, $f$ of $g$ of $z$.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this point in more detail, feel free to use the talk page.
If you are able to explain it, then when you have done so you can remove this instance of {{Explain}} from the code.
Proof
Follows directly from Complex Derivative of Product.
$\blacksquare$
