Combination Theorem for Complex Derivatives/Quotient Rule

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Theorem

Let $D$ be an open subset of the set of complex numbers $\C$.

Let $f, g: D \to \C$ be complex-differentiable functions.


Let $\dfrac f g$ denote the pointwise quotient of the functions $f$ and $g$.


Then $\dfrac f g$ is complex-differentiable in $D \setminus \left\{ {x \in D: g \left({z}\right) = 0}\right\}$.

For all $z \in D$ with $g \left({z}\right) \ne 0$:

$\left({\dfrac f g}\right)' \left({z}\right) = \dfrac{ f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{ \left({g \left({z}\right) }\right)^2 }$

where $\left({\dfrac f g}\right)'$ denotes the derivative of $\dfrac f g$.


Proof

Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $B_r \left({0}\right)$.

Let $z \in D \setminus \left\{ {x \in D: g \left({z}\right) = 0}\right\}$.

By the Alternative Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in B_{r_0} \left({0}\right) \setminus \left\{ {0}\right\}$:

$f\left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
$g\left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$

where $\epsilon_f, \epsilon_g: B_{r_0} \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$.

From Complex-Differentiable Function is Continuous, it follows that $g$ is continuous at $z$.

Then there exists $r_1 \in \R_{>0}$ such that for all $h \in B_{r_1} \left({0}\right)$, we have $\left\vert{g \left({z + h}\right) - g \left({z}\right) }\right\vert < \left\vert{g \left({z}\right) }\right\vert$.

Then by Backwards Form of Triangle Inequality:

$g \left({z + h}\right) \ne 0$ for all $h \in B_{r_1} \left({0}\right)$

Put $r = \min \left({r_0, r_1}\right)$.

Then for all $h \in B_r \setminus \left\{ {0}\right\}$:

\(\displaystyle \dfrac{f \left({z+h}\right) }{g \left({z+h}\right) }\) \(=\) \(\displaystyle \dfrac{f \left({z}\right) }{g \left({z}\right) } + \dfrac{f \left({z+h}\right) - f \left({z}\right) }{g \left({z+h}\right) } - f \left({z}\right) \dfrac{g \left({z+h}\right) - g \left({z}\right) }{g \left({z}\right) g \left({z+h}\right) }\) adding and subtracting the same quantity
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{f \left({z}\right) }{g \left({z}\right) } + \dfrac{h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) }{g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) } - \dfrac{ f \left({z}\right) h \left({ g' \left({z}\right) \epsilon_g \left({h}\right) }\right) }{g \left({z}\right) \left({g \left({z}\right) + h \left({ g' \left({z}\right) + \epsilon_g \left({h}\right) }\right) }\right) }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{f \left({z}\right) }{g \left({z}\right) } + h \dfrac{f' \left({z}\right) g \left({z}\right) + g \left({z}\right) \epsilon_f \left({h}\right) - f \left({z}\right) g' \left({z}\right) - f \left({z}\right) \epsilon_g \left({h}\right) }{ \left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) }\) with $\epsilon_0 \left({h}\right) = h g' \left({z}\right) \left({ g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{f \left({z}\right) }{g \left({z}\right) } + h\left({ \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{ {\left({g \left({z}\right) }\right)^2 } } + \epsilon \left({h}\right) }\right)\)

where $\epsilon: B_r \left({0}\right) \setminus 0 \to \C$ is defined by:

$\epsilon \left({h}\right) = \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) } - \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 } + \dfrac{ g \left({z}\right) \epsilon_f \left({h}\right) - f \left({z}\right) \epsilon_g \left({h}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) }$

From Combination Theorem for Continuous Functions, it follows that $\epsilon$ is continuous.

From Combination Theorem for Limits of Functions, it follows that:

\(\displaystyle \lim_{h \to 0} \epsilon \left({h}\right)\) \(=\) \(\displaystyle \lim_{h \to 0} \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) } - \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 } + \lim_{h \to 0} \dfrac{ g \left({z}\right) \epsilon_f \left({h}\right) - f \left({z}\right) \epsilon_g \left({h}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 } - \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 } + g \left({z}\right) \lim_{h \to 0} \dfrac{\epsilon_f \left({h}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right)} - f \left({z}\right) \lim_{h \to 0} \dfrac{\epsilon_g \left({h}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


Then the Alternative Differentiability Condition shows that:

$\left({\dfrac f g}\right)' \left({z}\right) = \dfrac {f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right)} {\left({g \left({z}\right) }\right)^2}$

$\blacksquare$


Sources