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I'm currently taking a course in mathemathical tools, where we are covering complex analysis (Note that this course is not very rigorous and we cover complex analysis in only 4 lectures). The Laurent series have been introduced as:

Laurent series$$ \begin{array}{l} f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n} \\ a_{n}=\frac{1}{2 \pi i} \oint_{C} \frac{f\left(z^{\prime}\right) d z^{\prime}}{\left(z^{\prime}-z_{0}\right)^{n+1}} \end{array} $$

However, I noticed that the above integral isn't used in the solutions to any of the problems regarding laurent series. So I'm trying to understand why we can avoid using it. What I understood so far, is that the difference between the Taylor and Laurent series, is that the Laurent series also contains negative powers. Where it's not possible to expand a Taylor series around a point $z_0$ where f is not analytically, the opposite applies for the Laurent series. In the case we expand f around a point where f is analytically, the Laurent series and Taylor expansion will be the same. If that is the correct understanding, please help me understand the solution to the following problem:


Find the Laurent series for $f(z)=(1-z)e^{1/z}$ about $z=0$

What I thought should be my approach is using the formula above solving the contour integral, since $f(z)$ is non-analytically at $z=0$. However, the solution uses directly that $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$. So here comes my first question:

  1. Why is the Laurent series of $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$.
    It seems to me, that they have just substituted $z \rightarrow 1/z$ in the Taylor series for $e^z$. But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist? And if that is not the Taylor series, how do we know it's the Laurent series?

After this they multiply the two expressions together: $(z-1) e^{1 / z}=(z-1) \sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=z-\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right) \frac{z^{-n}}{n !}$

  1. Is that because $z-1$ is a polynomial and thus is analytically around $z=0$ and is it's own Taylor series and Laurent series. So we can just find the Laurent series of each factor and multiply them together to find the final Laurent series?

Fx: Consider $h(z)=f(z)g(z)$, if we want to find the Laurent series around $z=z_0$. If the Laurent series of f(z) around $z_0$ is $\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}$ and of $g(z)$ is $\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n}$. Is the Laurent series of $h(z)$ then: $h(z)=(\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}) (\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n})$?

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But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist?

No. Not being analytic at point $z_0$ does not necessarily mean that Taylor expansion around $z_0$ does not exist. Consider the example $\frac{\sin z}{z}$ at $0$. The singularity is removable. Though $z=0$ is not a removable singularity of $e^{1/z}$.

On the other hand, you are looking for the Laurent series of $e^{1/z}$, not the "Taylor series".

All you do in this case is to look at the Taylor series (or definition, depending on how you define the exponential function) of $e^w$ at $w=0$: $$ e^w=\sum_{n=0}^\infty \frac{w^n}{n!}\tag{1} $$ This expansion is valide for any $w\in\mathbb{C}$. So you can do the substitution $w=\frac{1}{z}$ for any $z\ne 0$.

How do you know this is the Laurent series in your definition? The substitution above gives you the coefficients $a_n$. You can verify that $$ a_n=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z^{n+1}}dz $$ with $f(z)=e^{1/z}$.

Or, you can simply use the uniqueness of the coefficients.

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Think about what finding Laurent series of $g(z):=(1-z)e^{1/z}$ about $z=0$ really means. What you really looking for is a double-sided series $$ \sum_{n=-\infty}^\infty a_nz^n\tag{2} $$ such that at every $z$ with $0<|z|<R$ (for some $R$), (2) is equal to $g(z)$.

On the one hand, you can use your integral definition to find the coefficients. On the other hand, you can directly find the coefficients so that $$ g(z)=\sum_{n=-\infty}^\infty a_nz^n $$ at every $z$ with $0<|z|<R$. Since the coefficients of the Laurent series are unique, you would have the same answer no matter which way you take.

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Since, for each $z\in\Bbb C\setminus\{0\}$,$$e^{1/z}=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots,$$you have (again, for each $z\in\Bbb C\setminus\{0\}$),\begin{align}(1-z)e^{1/z}&=-z+(1-1)+\left(1-\frac1{2!}\right)\frac1z+\left(\frac1{2!}-\frac1{3!}\right)\frac1{z^2}+\left(\frac1{3!}-\frac1{4!}\right)\frac1{z^4}+\cdots\\&=-z+\sum_{n=0}^\infty\left(\frac1{n!}-\frac1{(n+1)!}\right)\frac1{z^n}.\end{align}


Concerning your final question: $\displaystyle\left(\sum_{n=-\infty}^\infty a_n\left(z-z_{0}\right)^{n}\right)\left(\sum_{n=-\infty}^\infty b_n\left(z-z_{0}\right)^{n}\right)$ is not a Laurent series.

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  • $\begingroup$ But how do I know that is the laurent series we have arrived at? It makes sense to me that $e^{1 / z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$ on $z \in \mathbb{C} \setminus \{0\}$, since we are just substituting into an already known equality. But is $e^{1 / z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$ a taylor expansion of $e^{1/z}$ around z=0, because that shouldn't exist (Isn't it rather an expansion around z=infinity considering the substitution, if that even makes sense?). And why would that be a laurent series? $\endgroup$
    – sjm23
    Feb 19 at 13:18
  • $\begingroup$ If $z_0$ is an isolated singularity of an analytic function, then there is one and only one Laurent series which converges to $f$ on a disk $D_r(z_0)$, which is the Laurent series of $f$. So, the way by which the Laurent series is obtained is nor relevant, as long as you get some Laurent series converging to $f(z)$. Just like you deduce from the equality$$|z|<1\implies\frac1{1-z}=\sum_{n=0}^\infty z^n$$that the power series $\sum_{n=0}^\infty z^n$ is the Taylor series of $\frac1{1-z}$ in the neighborhood of $0$. $\endgroup$ Feb 19 at 13:23
  • $\begingroup$ Concerning the second question. If that's not the case. What is it we are exploiting? How do we know that one function multiplied with a series gives the laurent series. Why is the function of $1-z$ multiplied with the series of $e^{1/z}$ necessarily a laurent series. $\endgroup$
    – sjm23
    Feb 19 at 13:23
  • $\begingroup$ I think that I have already provided an answer to that question. $\endgroup$ Feb 19 at 13:24
  • $\begingroup$ Okay, so the laurent series is unique in the way, that if I can express the function on the form of the equation of a laurent series (the equality holds using known expressions), then I know that has to be it's laurent series? $\endgroup$
    – sjm23
    Feb 19 at 13:28
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Let us first briefly try to understand the differences between the Laurent and the Taylor series. You have correctly observed that the Taylor series is only defined in a region in which the function is analytic. To be a little bit more specific: If $f$ is analytic at $z_0$, its Taylor series converges uniformly in the biggest circle centered around $z_0$ in which $f$ is analytic, i.e. the radius of the circle is the distance to the closest singularity. The Laurent series is however define around points were $f$ need not be analytic. In fact, we are interested in Laurent series presicely because they have this property. So suppose $f$ has an isolated singularity at $z_0$, then we can expand $f$ in a Laurent series which converges uniformly on a punctured circle around $z_0$ or an annulus. As you have observed the most apparent difference between the two series is that the Laurent series contains negative powers. This really means that is the sum of two series, one converging in an area $|z-z_0| < R$ and another convering in an area $|z-z_0| > r$. The sum will converge in the annulus $r < |z-z_0| < R$. Intuitively, the negative powers "make up for" the behaviour of $f$ close to the singularity.

However, if $f$ was actually analytic at $z_0$, the Laurent series will equal the Taylor series. You can easily see this by studying your expression for $a_n$. If $f$ is analytic at $z_0$ this equal $0$ for all $n<0$.

Now, to the computational aspect of your question. We usually ever use the integral formula for $a_n$, simply because it is tedious and unnecessary. It is used in proofs, but rarely to find Laurent expansions. It is a legitimate question why you can simply substitute $1/z$ for $z$ in the Taylor series of $e^z$. Think of it this way: Let $w \in \mathbb{C}$. You know that $e^w = \sum_{k=0}^{\infty} \frac{w^k}{k!}$. Now, let $z \neq 0$. $\frac{1}{z}$ is a complex number, so substituting $w = \frac{1}{z}$ is completely legitimate, and we get

$$e^{\frac{1}{z}} = \sum_{k=0}^{\infty} \frac{1}{z^k k!}.$$

Notice that this works whever $z \neq 0$, and that the series has now turned into a Laurent series. We claim that it converges for all $z \neq 0$, which follows immediately from how we constructed the series. The "annulus" in which the series converges is $\{ z \in \mathbb{C} \, | \, |z| > 0 \} = \mathbb{C} \setminus \{ 0 \}$. Using this we can easily find the Laurent series you are asked to find:

$$(1-z)e^{1/z}=e^{1/z}-ze^{1/z}=\sum_{k=0}^{\infty} \frac{1}{z^k k!}-z \sum_{k=0}^{\infty} \frac{1}{z^k k!} = \sum_{k=0}^{\infty} \frac{1}{z^k k!} - \sum_{k=0}^{\infty} \frac{1}{z^{k-1} k!}$$

Rearranging this, we get

$$(1-z)e^{1/z} = -z + \sum_{k=0}^{\infty} \bigg(\frac{1}{k!}-\frac{1}{(k+1)!} \bigg) \frac{1}{z^k}.$$

At last, how do we know that this is the Laurent series and not some other series converging to $(1-z)e^{1/z}$ ? Well, this is a theorem one would usually prove in an introductory course in complex analysis. A series representation of a function of the form

$$\sum_{n=-\infty}^{\infty} a_n z^n$$

is unique.

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