# Laurent Series, poles and zeros

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 |

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 $$(z-1/2)$$.

• 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 $a (z-\frac 12) + \frac 12 a + b + \frac {c}{2z + 1}$ Commented Jan 31, 2020 at 18:45
• Thanks for your answer, was really helpful @Doug M Commented Jan 31, 2020 at 19:29

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.

• 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. Commented Jan 31, 2020 at 19:19
• Very glad to help! Commented Feb 1, 2020 at 5:43

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

• I didn't even realize that I could have done that, thank you for your answer @Saketh Malyala Commented Jan 31, 2020 at 19:17

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 :-).

• 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 Commented Jan 31, 2020 at 19:24