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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?

Thanks in advance!

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  • $\begingroup$ 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 $f$." $\endgroup$ Commented Oct 7, 2019 at 16:37
  • $\begingroup$ @Barry Cipra, 8th edition and page 249 of section 75 zeros of analytic function. $\endgroup$
    – anadad
    Commented Oct 7, 2019 at 16:44
  • $\begingroup$ 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. $\endgroup$
    – anadad
    Commented Oct 7, 2019 at 16:46
  • $\begingroup$ Oh. It says about a zero of order... I’m sorry! But I believe I never seen a definition of zero in other pages. $\endgroup$
    – anadad
    Commented Oct 7, 2019 at 16:48
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    $\begingroup$ 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. $\endgroup$ Commented Oct 7, 2019 at 17:08

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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.

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    $\begingroup$ 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? $\endgroup$
    – anadad
    Commented Oct 7, 2019 at 17:40
  • $\begingroup$ 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 $f$. $\endgroup$ Commented Oct 7, 2019 at 17:49
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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.

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