# understanding the difference between Laurent and Taylor series.

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?

• Laurent series could have terms of negative degree – 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.

• 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? – Qhef Apr 14 '19 at 21:52
• 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 $0$. – José Carlos Santos Apr 14 '19 at 21:57
• @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 $\infty$. – lc2r43 Apr 14 '19 at 22:03
• 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? – Qhef Apr 14 '19 at 22:16
• Right. It is $0$ for that reason. – 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.