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I'm trying to figure out problem b from this set of homework questions:

Find the Laurent series expansions of $$ \textrm{(a) } \frac{e^z}{z^4},\qquad \textrm{(b) } ze^{1/z},\qquad \textrm{(c) } \frac{1}{z(z-1)} $$

The expansion is taken around zero.

Currently it's my understanding that you take a Taylor expansion of a part of the function that is analytic and then multiply by the bit that remains. However for question b my lecturer has taken a Taylor expansion for $e^{1/z}$. My confusion is that isn't $z = 0$ for $e^{1/z}$ a singular point and therefore he can't take a Taylor expansion of it and simply multiple by $z$ to get his answer?

Any help would be appreciated

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  • $\begingroup$ You seriously need to typeset, now here at MSE, those questions from the link. Tha link, after all, can easily disappear over time. $\endgroup$
    – zhw.
    Commented Feb 22, 2017 at 22:41
  • $\begingroup$ So how do I show the problem as you did without a link? $\endgroup$ Commented Feb 23, 2017 at 10:00
  • $\begingroup$ oh I see how you have done that $\endgroup$ Commented Feb 23, 2017 at 10:01

2 Answers 2

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$e^{\frac{1}{z}}$ does have an essential singularity at zero, it doesn't have a Taylor series, but it does have a Laurent expansion. $$e^{\frac{1}{z}}=\cdots \frac{1}{2!z^2}+\frac{1}{z} +1$$

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  • $\begingroup$ can you tell me how you derived that laurent expansion? $\endgroup$ Commented Feb 22, 2017 at 22:25
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    $\begingroup$ I took the usual taylor series for $e^z$ and substituted $\frac{1}{z}$. $\endgroup$ Commented Feb 22, 2017 at 22:29
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You know that for all $w \in \mathbb C$ you have $$e^w=1+\frac{1}{1!}w+\frac{1}{2!}w^2+...$$

Now, if $z \in \mathbb C$ is non zero, setting $w=\frac{1}{z}$ you get $$e^{\frac{1}{z}}= 1+\frac{1}{1!z}+\frac{1}{2!z^2}+...$$

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