A problem on meromorphic functions

Question

Consider the function $$F(z)=\int_{1}^{2}{1\over (x-z)^2}dx,\;\;\;\operatorname{Im}(z)>0$$

Then there is a meromorphic function $G(z)$ on $\mathbb{C}$ that agrees with $F(z)$ when $\operatorname{Im}(z)>0$ such that

Statement 1: $1$, $\infty$ are poles of $G(z)$.

Statement 2: $0$, $1$, $\infty$ are poles of $G(z)$.

Why both of these statement are false?

Kindly give me some hint, so that I can solve this problem.

Edits:

I have already solved the integral and found out that $z=1$ and $z=2$ are simple poles of $F(z)$ but I don't know whether I can use this information to show that there does not exist $G(z)$ with the above mentioned properties.

I solved the integral and found out that $F(z)$ has simple poles at $1 $and $2$ @LordSharktheUnknown — Shweta Aggrawal, Oct 21 '18 at 12:04
But how does that prove that there does not exist a Meromorphic function $G$ with these properties. If I take $G(z)=F(z)$ I can only show that there exist $G$ with poles at $z=1$ and $z=2$ @LordSharktheUnknown — Shweta Aggrawal, Oct 21 '18 at 12:07
@LordSharktheUnknown $$\int_1^2 \frac{1}{(x-z)^2},dx = \biggl[\frac{1}{z-x}\biggr]_1^2 = \frac{1}{z-2} - \frac{1}{z-1}$$ — Shweta Aggrawal, Oct 21 '18 at 12:35
Excellent; that looks like a meromorphic function. Where's its poles? — Angina Seng, Oct 21 '18 at 13:09
@LordSharktheUnknown Poles are at $z=1$ and $z=2$, But $\frac{1}{z-2} - \frac{1}{z-1}$ is equal to $F(z)$. How does knowing what $F$ is helps me in deducing result about $G$. I understand that as a particular case I can take $G=F$ but can not be there other choices for $G$. — Shweta Aggrawal, Oct 21 '18 at 13:12

score 0 · Accepted Answer · answered Oct 23 '18 at 14:52

Let us determine what $F(z)$ is by evaluating the integral,

$$F(z)=\int_1^2 \frac{1}{(x-z)^2}\,dx = \biggl[\frac{1}{z-x}\biggr]_1^2 = \frac{1}{z-2} - \frac{1}{z-1}$$

Let $G$ be a meromorphic function which agrees with $F(z)$ on the given domain.

$F$ has simple poles at $z=1, z=2$ and in particular we can take $G(z)=F(z)$ so that means $G(z)$ also has simple poles at $z=1,z=2$.

Now by a theorem in complex analysis, which roughly says that there is a unique meromorphic function having given set of poles.

We conclude that any such $G$ will have poles at $z1,z=2$. Therefore these statements are false.

A problem on meromorphic functions

1 Answers1