# A definition of zeros of a complex function.

I’m studying complex variables with a book written by Churchill.

In the textbook, he defines a point as a zero of a function only when it is zero and analytic at the point. But I don’t understand why we need analyticity when we talk about zeros.

I suspect that people didn’t want a point to be zeros and poles of a function at the same time. If we didn’t require a function to be analytic, then, for example, we just can bring a function which has a pole at a point z and redefine the function at a point z by zero. Then z is a zero and a pole at the same time. (* A pole of a function is an isolated singular point at which the principal part of its Laurent Series has at least one non-zero term and finitely many non-zero terms)

But I don’t know why they shouldn’t appear at the same point. Is there any reason for this?

• Can you specify the edition and page number for this definition, and perhaps quote it verbatim? I am looking at a second edition of Churchill (from 1960), and on page 52 it simply says "A value of $z$ for which $f(z)=0$ is called a zero of the function $f$." Commented Oct 7, 2019 at 16:37
• @Barry Cipra, 8th edition and page 249 of section 75 zeros of analytic function. Commented Oct 7, 2019 at 16:44
• Suppose that f is analytic at z. If f(z) = 0 and if there is a positive integer m such that m-th derivative at z is not zero and each derivative of lower order vanishes at z , then f is said to have a zero of order m at z. Commented Oct 7, 2019 at 16:46
• Oh. It says about a zero of order... I’m sorry! But I believe I never seen a definition of zero in other pages. Commented Oct 7, 2019 at 16:48
• Poles and zeros are quite different, and a pole is not a zero. What is the case is that $f$ has a pole of order $m$ at a point $z$ if and only if its reciprocal, $1/f$, is analytic at $z$ and has a zero of order $m$ there. Being analytic is quite important here. Commented Oct 7, 2019 at 17:08

The condition of being analytic for $$f(z)$$ at its zero $$z=a$$(say) is inherent which can be verified by just looking at the definition of the order of a zero of $$f(z)$$ i.e.

"A function $$f(z)$$ has a zero of order $$m$$ at $$z=a$$ if $$f(z)=(z-a)^mg(z)$$, where $$g(a)\ne 0$$ and $$g(z)$$ is analytic at $$z=a$$ ".

Analyticity of $$g$$ at $$z=a$$ necessitates analyticity of $$f$$ at $$z=a$$. In contrast, while displaying the functional value of a real function $$f(x)$$ at a point $$x=a$$ of its domain, it is not necessary that $$f$$ is continuous/differentiable there. For example, $$f(x)=0$$, if $$x\le 0$$ and $$f(x)=1$$ elsewhere on $$\mathbb R$$, you can see $$f(0)=0$$. Further, $$x=a$$ is not necessarily the limit point of the domain of $$f$$ for defining $$f(a)$$ e.g. if $$A=$${$$0$$}$$\cup (1,\infty)$$ and a function $$f:A\rightarrow \mathbb R$$ defined as $$f(x)=0$$ if $$x=0$$ and $$f(x)=x$$ elsewhere is a continuous function on its domain. However, the domain of a complex continuous function can't have isolated points inside it.

• Ah that makes sense. One more question! If we define a function f by &f(1)=0 and f(z)=1/(z-1)& Then z=1 is clearly a pole of order 1. Is it a zero? Commented Oct 7, 2019 at 17:40
• Defining $f$ in such a way leads to a contradiction at $z=1$ actually. A point $z=a$ is either a singularity of $f$ or a point in domain of analyticity of $f$. Commented Oct 7, 2019 at 17:49

The function

$$f(z)=\begin{cases}{1\over z}&\text{if }z\not=0\\0&\text{if }z=0\end{cases}$$

(similar to what the OP asks about in comments) meets the definitions in (my edition, at least, of) Churchill for having both a zero at $$z=0$$ and a (simple) pole at $$z=0$$. In particular, the definition of a zero does not mention anything about continuity (much less analyticity), while the definitions of (isolated) singularities and poles do not mention whether the function is or is not assigned a value at the singular point, only that it's analytic in a neighborhood around the point.

So technically yes, a function can have both a zero and a pole at the same point. This is kind of a trick answer to a trick question: the assignment of the value $$0$$ to $$f(0)$$ is quite arbitrary, since no value makes the continuous at $$z=0$$. But kudos to the OP for wondering whether Churchill's definitions allow it.