Analyticity: Sums, Differences, Products, Quotients of Analytic Functs.

# Analyticity of Sums, Differences, Products, and Quotients of Analytic Functions

Recall from the Analytic Complex Functions page that if $A \subseteq \mathbb{C}$ is open, $z_0 \in A$, and $f : A \to \mathbb{C}$ then $f$ is said to be analytic at $z_0$ if there exists an $r > 0$ such that $f$ is complex differentiable on the open disk $D(z_0, r)$. Furthermore, we said that $f$ is analytic on $A$ if $f$ is analytic at every point $z_0 \in A$.

We will now look at some results regarding the analyticity of sums, differences, products, and quotients of analytic functions - all of which results are analogous to that of the corresponding theorems from calculus.

Theorem 1: Let $A \subseteq \mathbb{C}$ be an open set, $f, g : A \to \mathbb{C}$ be analytic on $A$, and let $k \in \mathbb{C}$. Then:a) $f + g$ is analytic on $A$ and $(f + g)'(z) = f'(z) + g'(z)$.b) $kf$ is analytic on $A$ and $(kf)'(z) = kf'(z)$.c) $fg$ is analytic on $A$ and $(fg)'(z) = f(z)g'(z) + f'(z)g(z)$.d) $\displaystyle{\frac{f}{g}}$ is analytic on $A$ and $\displaystyle{\left ( \frac{f}{g} \right )'(z) = \frac{f'(z)g(z) - f(z)g'(z)}{[g(z)]^2}}$ (provided that $g'(z) \neq 0$ for all $z \in A$). |

**Proof of a)**Let $f$ and $g$ be analytic on $A$. Then $f$ and $g$ are differentiable at every point $z_0 \in A$, so $f'(z_0)$ and $g'(z_0)$ exist and:

\begin{align} \quad (f + g)'(z_0) &= \lim_{z \to z_0} \frac{[f(z) + g(z)] - [f(z_0) + g(z_0)]}{z - z_0} \\ &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} + \lim_{z \to z_0} \frac{g(z) - g(z_0)}{z - z_0} \\ &= f'(z_0) + g'(z_0) \end{align}

- So $(f + g)'(z_0)$ exists for every $z_0 \in A$ which shows that $f + g$ is analytic on $A$ and that $(f + g)'(z) = f'(z) + g'(z)$. $\blacksquare$

**Proof of b)**Let $f$ be analytic on $A$ and let $k \in \mathbb{C}$. Then $f$ is differentiable at every point $z_0 \in A$, so $f'(z_0)$ exists and:

\begin{align} \quad (kf)'(z_0) &= \lim_{z \to z_0} \frac{kf(z) - kf(z_0)}{z - z_0} \\ &= k \cdot \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \\ &= k f'(z_0) \end{align}

- So $(kf)'(z_0)$ exists for every $z_0 \in A$ which shows that $kf$ is analytic on $A$ and that $(kf)'(z) = kf'(z)$. $\blacksquare$

**Proof of c)**Let $f$ and $g$ be analytic on $A$. Then $f$ and $g$ are differentiable at every point $z_0 \in A$, so $f'(z_0)$ and $g'(z_0)$ exist, and:

\begin{align} \quad (fg)'(z_0) &= \lim_{z \to z_0} \frac{f(z)g(z) - f(z_0)g(z_0)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{f(z)g(z) - f(z)g(z_0) + f(z)g(z_0) - f(z_0)g(z_0)}{z - z_0} \\ &= \lim_{z \to z_0} f(z) \frac{g(z) - g(z_0)}{z - z_0} + \lim_{z \to z_0} g(z_0) \frac{f(z) - f(z_0)}{z - z_0} \\ &= f(z_0) g'(z_0) + f'(z_0)g(z_0) \quad (\mathrm{By \: the \: continuity \: of \:} f) \end{align}

- So $(fg)'(z_0)$ exists for every $z_0 \in A$ which shows that $fg$ is analytic on $A$ and that $(fg)'(z) = f(z)g'(z) + f'(z)g(z)$. $\blacksquare$

**Proof of d)**Let $f$ and $g$ be analytic on $A$ with $g(z) \neq 0$ for all $z \in A$. Then $f$ and $g$ are differentiable at every point $z_0 \in A$, so $f'(z_0)$ and $g'(z_0)$ exist, and:

\begin{align} \quad \left ( \frac{f}{g} \right ) '(z_0) &= \lim_{z \to z_0} \frac{\frac{f(z)}{g(z)} - \frac{f(z_0)}{g(z_0)}}{z - z_0} \\ &= \lim_{z \to z_0} \frac{\frac{f(z)g(z_0) - f(z_0)g(z)}{g(z)g(z_0)} }{z - z_0} \\ &= \lim_{z \to z_0} \frac{1}{g(z)g(z_0)} \cdot \frac{f(z)g(z_0) - f(z_0)g(z)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{1}{g(z)g(z_0)} \cdot \frac{f(z)g(z_0) - f(z)g(z) + f(z)g(z) - f(z_0)g(z)}{z - z_0} \\ &= \lim_{z \to z_0} \frac{1}{g(z)g(z_0)} \cdot \left [ -f(z) \cdot \frac{g(z) - g(z_0)}{z - z_0} + g(z) \frac{f(z) - f(z_0)}{z - z_0} \right ] \\ &= \lim_{z \to z_0} \frac{f'(z)g(z) - f(z)g'(z)}{[g(z)]^2} \end{align}

- So $\displaystyle{\left ( \frac{f}{g} \right ) '(z_0)}$ exists for every $z_0 \in A$ which shows that $\displaystyle{\frac{f}{g}}$ is analytic on $A$ and that $\displaystyle{\left ( \frac{f}{g} \right )'(z) = \frac{f'(z)g(z) - f(z)g'(z)}{[g(z)]^2}}$. $\blacksquare$

Theorem 2: Let $A \subseteq \mathbb{C}$ be an open and connected and let $f : A \to \mathbb{C }$ be analytic on $A$. If $f'(z) = 0$ for all $z \in A$ then $f$ is constant on $A$. |

**Proof:**Let $f = u + iv$ and let $z_1, z_2 \in A$ be distinct points in $A$. Since $A$ is an open and connected it is path connected. So there exists a continuous differentiable function $\gamma$ in $A$ joining $z_1$ to $z_2$. Hence:

\begin{align} \quad \frac{d}{dt} f(\gamma(t)) = f'(\gamma(t)) \cdot \gamma'(t) \end{align}

- But $f'(\gamma(t)) = 0$ since $f'(z) = 0$ for all $z \in A$, i.e., $\frac{d}{dt} f(\gamma(t)) = 0$. Also:

\begin{align} \quad \frac{d}{dt} f(\gamma(t)) = \frac{d}{dt} u(\gamma(t)) + i \frac{d}{dt} v(\gamma(t)) = 0 \end{align}

- A complex number is zero if and only if its real and imaginary parts are both zero. Therefore $\frac{d}{dt} u(\gamma(t)) = 0$ and $\frac{d}{dt} v(\gamma(t)) = 0$. But $u$ and $v$ are real-valued functions so $u(\gamma(t))$ and $v(\gamma(t))$ must be constant. So $f(z_1) = f(z_2)$ and thus $f$ is constant on $A$. $\blacksquare$