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I've been having a bad time with the Laurent Series, so I would really appreciate if someone can help me with this problem:

Consider $f(z)$ $$ f(z) = \frac{z^2+1}{2z-1} $$ Then:

i) Classify the zeros and poles of $f(z)$, consider the extended complex plane

ii) Find an expansion in the Laurent Series of $f(z)$ valid in some region of the form $0 <| z-z_0 | <R$

ii) Based on what was obtained in the previous point, find the residue at $z_0$.

My English is not very good, so I apologize for the possible grammatical mistakes.

i) I found that the zeros of the function are $i$ and $-i$, both of order 1. Then, for the zero at infinity, I consider $g(z_1)$ $$ g(z_1) = f \left( \frac{1}{z_1}\right)=\frac {(1-z_1)(1+z_1)}{2-z_1} $$ and I evaluated it at $z_1 = 0$, and concluded that $f (z)$ does not have a zero at infinity. For the poles, I found that there is a simple pole at $z = 1/2$. Then I consider $g (z_1)$ and see if it has a pole at $z_1 = 0$, and I concluded that $f (z)$ does not have a pole at infinity.

ii) I think that I need to find the Laurent expansion around $z_0= 1/2$, but i dont know how to express $f(z)$ in terms of $(z-1/2)$.

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  • $\begingroup$ For ii) you could say $w = z - \frac 12 \implies z = w + \frac 12$ and substitute. Or you could do polynomial division on the numerator by the denominator, and you will get $az + b + \frac {c}{2z-1}$ to which you can say $a (z-\frac 12) + \frac 12 a + b + \frac {c}{2z + 1}$ $\endgroup$
    – Doug M
    Jan 31 '20 at 18:45
  • $\begingroup$ Thanks for your answer, was really helpful @Doug M $\endgroup$
    – Emiliano
    Jan 31 '20 at 19:29
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By means of long division you have that $$ f(z) = \frac{1} {2}z + \frac{1} {4} + \frac{5} {4}\frac{1} {{\left( {2z - 1} \right)}} $$ Now you can rewrite it as $$ f(z) = \frac{1} {2}\left( {z - \frac{1} {2} + \frac{1} {2}} \right) + \frac{1} {4} + \frac{5} {8}\frac{1} {{\left( {z - \frac{1} {2}} \right)}} $$ or $$ f(z) = \frac{1} {2}\left( {z - \frac{1} {2}} \right) + \frac{1} {2} + \frac{5} {8}\frac{1} {{\left( {z - \frac{1} {2}} \right)}} $$ which is the require Laurent development.

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  • $\begingroup$ In one moment I thought in do the division, but I didn't try it. Thanks for your answer, was really helpful @Luca Goldoni Ph.D. $\endgroup$
    – Emiliano
    Jan 31 '20 at 19:19
  • $\begingroup$ Very glad to help! $\endgroup$ Feb 1 '20 at 5:43
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It should be simple enough to find the Taylor Series of $z^2+1$ around $z=\frac{1}{2}$.

This would be $\displaystyle \frac{5}{4}+\left(z-\frac{1}{2}\right)+\left(z-\frac{1}{2}\right)^2$

Then, divide your polynomial by $2\displaystyle \left(z-\frac{1}{2}\right)$ (the denominator) to get your Laurent series.

You would then get $\displaystyle \frac{5}{8}\left(z-\frac{1}{2}\right)^{-1}+\frac{1}{2}+\frac{1}{2}\left(z-\frac{1}{2}\right)$

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    $\begingroup$ I didn't even realize that I could have done that, thank you for your answer @Saketh Malyala $\endgroup$
    – Emiliano
    Jan 31 '20 at 19:17
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First, note that $f({1 \over z}) = - {1 +z^2 \over z(z-2)}$ which has a pole at $ z=0$ so $f$ does have a pole at infinity.

You can expand $f$ anywhere, but it is fairly easy to expand around the pole $z_0={1 \over 2}$. To simplify life, consider the mapping $w=2z-1$ which gives $z={1 \over 2}(w+1)$ and define $\phi(w) = f({1 \over 2}(w+1))={5 \over 4} {1 \over w} + {1 \over 2} + {1 \over 4} w$ whose Laurent expansion around $w=0$ is clear. Now map back to the '$z$' variable: $f(z) = \phi(2z-1) = {5 \over 8} {1 \over z-{1 \over 2}} + {1 \over 2} + {1 \over 2} ( z-{1 \over 2})$. You can read off the residue from the $f_{-1}$ term as ${5 \over 8}$. This series is valid for all $0<|z-{1 \over 2}|$, so $R=\infty$.

If you wanted to expand $f$ around some $z_0 \neq {1 \over 2}$, first note that the expansion can be at most valid for $0< |z-z_0| < |{1 \over 2}-z_0|$. Since $f$ is analytic in this neighbourhood, we expect a power series expansion (that is, all $f_{-n}$ coefficients are zero). Expanding about $z=z0$ we have \begin{eqnarray} f(z) &=& {z^2+1 \over 2z-1} \\ &=& {z^2+1 \over 2(z-z_0)-(1-2z_0)} \\ &=& {1 \over 1-2z_0} {z^2+1 \over {2(z-z_0) \over 1-2 z_0} -1} \\ &=& {1 \over 1-2z_0} (z^2+1) \sum_{n=0}^\infty ({2(z-z_0) \over 1-2 z_0})^n\\ &=& {1 \over 1-2z_0} (z^2+1) \sum_{n=0}^\infty ({2 \over 1-2 z_0})^n(z-z_0)^n\\ &=& {1 \over 1-2z_0} ((z-z_0+z_0)^2+1) \sum_{n=0}^\infty ({2 \over 1-2 z_0})^n(z-z_0)^n\\ &=& {1 \over 1-2z_0} ((z-z_0)^2+2(z-z_0)+1) \sum_{n=0}^\infty ({2 \over 1-2 z_0})^n(z-z_0)^n\\ \end{eqnarray} from which we can read off the power series coefficients (in particular $f_{-1} = 0$).

However, it is clear that expanding around the pole uses far fewer symbols :-).

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  • $\begingroup$ I saw your first comment about the pole, then I checked it and realized where I was wrong, thanks. And this method would probably never have crossed my mind, thank you very much @copper.hat $\endgroup$
    – Emiliano
    Jan 31 '20 at 19:24

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