Big-oh notation for complex functions.

Question

I can not find the big-oh notation for complex fuctions, even though it should be pretty clear.

Lets say that it is said $$f(z)=\mathcal{O(g(z))}, z \rightarrow \infty,$$

where there may be complex values where $f$ is not defined. Can we just assume that this means that there exists positive real numbers $M,N$ such that

$$|f(z)|\le N|g(z)|$$

if $|z| > M$, and $f(z),g(z)$ are defined?

Is the big-oh notation for complex functions defined in any books you know?

Answer 1

Here are some references which might be helpful.

• From Asymptotics and Special Functions by F.W.J. Olver, ch. 1, section 3, 3.2:

  The next extension is to complex variables. Let $S$ be a given infinite sector $\alpha \leq \text{ph }z\leq \beta$ denoting the phase or argument of $z$. Suppose that for a certain value of $R$ there exists a number $K$, independent of $\text{ph }z$, such that \begin{align*} |f(z)|\leq K|\Phi(z)|\qquad\qquad (z\in S(R)) \end{align*} where $S(R)$ denotes the intersection of $S$ with the annulus $|z|\geq R$. Then we say that $f(z)=O\{\Phi(z)\}$ as $z\rightarrow \infty$ in $S$, or, equivalently, $f(z)=O\{\phi(z)\}$ in $S(R)$. Thus the symbol $O$ automatically implies uniformity with respect to $\text{ph }z$.

• From Asymptotic Methods in Analysis by N.G. De Bruijn, ch. 1, section 2:

  ... the $O$-symbol is used in the sense of something that is, in absolute value, at most a constant multiple of the absolute value of. So if $S$ is any set, and if $f$ and $\phi$ are real or complex functions defined on $S$, then the formula \begin{align*} f(s)=O(\phi(s))\qquad\qquad (s\in S) \end{align*} means that there is a positive number $A$, not depending on $s$, such that \begin{align*} |f(s)|\leq A|\phi(s)|\qquad\qquad\text{for all }s\in S. \end{align*}

• From Analytic Combinatorics by P. Flajolet and R. Sedgewick, appendix A.2:

  Let $\mathbb{S}$ be a set and $s_0\in \mathbb{S}$ a particular element of $S$. We assume a notion of neighbourhood to exist on $\mathbb{S}$. Examples are $\mathbb{S}=\mathbb{Z}_{>0}\cup\{+\infty\}$ with $s_0=+\infty$, $\mathbb{S}=\mathbb{R}$ with $s_0$ any point in $\mathbb{R}$; $\mathbb{S}=\mathbb{C}$ or a subset of $\mathbb{C}$ with $s_0=0$, and so on. Two functions $\phi$ and $g$ from $\mathbb{S}\setminus\{s_0\}$ to $\mathbb{R}$ or $\mathbb{C}$ are given. Write \begin{align*} \phi(s){=_{s\rightarrow s_0}}O(g(s)) \end{align*} if the ratio $\phi(s)/g(s)$ stays bounded as $s\rightarrow s_0$ in $\mathbb{S}$. In other words, there exists a neighbourhood $V$ of $s_0$ and a constant $C>0$ such that \begin{align*} |\phi(s)|\leq C|g(s)|,\qquad\qquad s\in V, s\neq s_0. \end{align*}

  One also says that $\phi$ is of order at most $g$, or $\phi$ is big-Oh of $g$ (as $s$ tends to $s_0$).

N.G. Bruijn also states in section 1.2: There are some minor differences between the various definitions of the $O$-symbol that occur in the literature, but these differences are unimportant.

Answer 2

Complex numbers can't be compared and one needs to switch to some scalar anyway. Then there is no real need to introduce a new concept. You can write

$$|f(z)|=\mathcal{O(g(z))}$$ where $g$ is a real function (possibly $|g(z)|$ for some complex function $g$). Or use another property than the modulus.