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I am solving the following Exam question:

Question Choose the correct options from the following

Consider the function $\displaystyle{% \,\mathrm{F}\left(\, z\,\right) = \int_{1}^{2}{\mathrm{d}x \over \left(x - z\right)^{2}} \,,\quad \Im\left(\, z\,\right) > 0} $

Then there is a mermorphic function $G(z)$ on $\mathbb{C}$ that agrees with $\,\mathrm{F}(z)$ when $\Im\left(z\right)>0$, such that

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

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

  3. $1$, $2$ are poles of $G(z)$.

  4. $1$, $2$ are simple poles of $G(z)$.

My approach
As question says Meromorphic function $G(z)$ agrees with $\,\mathrm{F}(z)$ on $\Im(z)>0$. We can take $G(z)= \int_{1}^2\frac{1}{\left(\, x - z\,\right)^{\,\, 2}}\,\mathrm{d}x$. If we integrate this function with respect to $x$ assuming $z$ as a constant we will get $G(z) = {1 \over 2 - z} - {1 \over 1-z}$. This shows that given function $G(z)$ has simple poles of order $1$ and $2$. Hence options $3$ and $4$ are correct.

My doubt. I am not sure whether my approach is correct as the way question has been projected as an integration of a complex function where variable of integration is $x$ is new to me. Also, whats the logic behind considering $\Im(z)>0$?

Thank you for your time.

Felix Marin
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Srijan
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1 Answers1

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Your approach is essentially correct, but you have a sign error. $z$ is the parameter of the integral, so we need the primitive of $\frac{1}{(x-z)^2}$ with respect to $x$, and get

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

for $\operatorname{Im} z > 0$ (since only such $z$ are considered by definition).

This clearly extends to a meromorphic function on the whole plane, with (simple) poles at $1$ and $2$ and nowhere else.

The restriction $\operatorname{Im} z > 0$ ensures that the integral exists. We could also ensure that by allowing $z \in \mathbb{C}\setminus [1,2]$, but if $z\in [1,2]$, the integrand has a non-integrable singularity, so we must forbid those in the integral, although $F$ extends holomorphically to $(1,2)$.

Daniel Fischer
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  • Than you very much for the answer. It is beautifully explained. What I understood from your answer is: By considering $im(z)>0$, question excluded the possibility of $z$ being equal to $2$ or $1$. Thanks again. – Srijan Nov 17 '16 at 00:35
  • Not only $1$ or $2$, we must also exclude the numbers in between, like $\sqrt{2}$ or $\frac{8}{5}$. For such numbers, the integral doesn't exist. But if one stays away from the interval $[1,2]$, all works well. – Daniel Fischer Nov 17 '16 at 00:37
  • Really glad to discuss with you. Thanks a ton. :) – Srijan Nov 17 '16 at 00:39