In my homework, I have a problem that says,

Set $f(z)$ = $\frac{e^{z^2}}{z^4}$.

$(a):$ Find the Laurent series for $f$ centered at $z_0 = 0$
$(b):$ Let $C$ be the positively oriented unit circle. Evaluate $\int_C f(z) dz$

I don't think I understand how to find a Laurent series, or rather I don't understand the difference between finding Laurent series and Taylor series.

For part a, I said

"Recall that the Taylor series at $z_0 = 0$ of $e^z$ is $\sum_{k=0}^\infty$ $\frac{z^k}{k!}$. So the Taylor series of $e^{z^2}$ is just $\sum_{k=0}^\infty$ $\frac{z^{2k}}{k!}$. Therefore, the Taylor series expansion of $f(z)$ is just $\sum_{k=0}^\infty$ $\frac{z^{2k}}{z^4 k!}$."

When I asked Wolfram alpha what the Taylor series was, it spat out this exact series and told me it was a Laurent series. Is it a Laurent series? If so, why does its Taylor series equal its Laurent series?

  • $\begingroup$ Laurent series could have terms of negative degree $\endgroup$ – J. W. Tanner Apr 14 '19 at 21:46

No, they're not equal. The Taylor series of $e^{z^2}$ is$$\sum_{k=0}^\infty\frac{z^{2k}}{k!}=1+z^2+\frac{z^4}{2}+\frac{z^6}{3!}+\cdots$$The Laurent series of $\dfrac{e^{z^2}}{z^4}$ is$$\sum_{k=0}^\infty\frac{z^{2k}}{z^4k!}=\frac1{z^4}+\frac1{z^2}+\frac12+\frac{z^2}{3!}+\cdots$$They are distinct.

  • $\begingroup$ Forgive me if I'm misinterpreting this, Does that mean that $\frac{e^{z^2}}{z^4}$ doesn't even have a Taylor series representation? Essentially, it can only be represented as a Laurent series? $\endgroup$ – Qhef Apr 14 '19 at 21:52
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    $\begingroup$ The function $\frac{e^{z^2}}{z^4}$ isn't even defined at $0$, and therefore it cannot possibly have a Taylor series centered at $0$. $\endgroup$ – José Carlos Santos Apr 14 '19 at 21:57
  • $\begingroup$ @Qhef You are correct that around the point $z=0$, $\frac{e^{z^2}}{z^4}$ does not have a Taylor series, as it is not analytic at 0 (from the Laurent series we can see 0 is a fourth order pole). Of course, 0 (and the essential singularity at $\infty$) are the only points around which the given function is non-analytic, so it is possible to find a Taylor series centered at any point other than 0 or $\infty$. $\endgroup$ – lc2r43 Apr 14 '19 at 22:03
  • $\begingroup$ So then, since this is THE Laurent series for $f(z)$, we can see that there is no $1/z$ term and therefore the residue of the function is just 0 and since $\int_C f(z) dz$ = $ 2 \pi i (\sum{residues})$, the integral of the function about C is just 0? $\endgroup$ – Qhef Apr 14 '19 at 22:16
  • $\begingroup$ Right. It is $0$ for that reason. $\endgroup$ – José Carlos Santos Apr 14 '19 at 22:18

A Laurent series is allowed to (but does not have to) contain terms with negative exponents.

If it does contain such terms, then either function it describes has a pole at the point we're developing the series around, and therefore it does not have a Taylor series, or the Laurent series does not converge close to $z_0$.

If it does have a Taylor series around a point, that series is also its Laurent series around that point.


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