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:
- 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:
$(z-1) e^{1 / z}=(z-1) \sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=z-\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right) \frac{z^{-n}}{n !}$
- 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:
$h(z)=(\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}) (\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n})$?
Best Answer
1
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.
2
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.