# Laurent Series of $(1+z)e^{ \frac{1}{z}}$ around $z_0 = 0$

I am trying to find the Laurent Series of $f(z) = (1+z)e^{ \frac{1}{z} }$ around $z_0 = 0$, but I feel that I am not doing something right. I used the Taylor series of $e^z$ and then replaced the $z$ with $\frac{1}{z}$ and then tried multiplying it with the $(1+z)$ term. So, I have $(1+z)(\frac{1}{z} + \frac{1}{2z^2} + \frac{1}{6z^3})$, but find that it is not actually finding the Laurent series. Can somebody please explain what I am doing wrong and guide me through the problem?

I am using the Complex Analysis, Third Edition textbook by Joseph Bak and Donald J. Newman.

First, compute the Laurent series of $e^{1/z}$. We do this by taking the Taylor expansion of $e^z$ about $z=0$, assuming it's ok to replace $z$ with $1/z$, and then we use uniqueness of the Laurent series:

$$e^z = 1+z+\frac{z^2}{2!}+\cdots$$

So

$$e^{1/z} = 1+z^{-1}+\frac{z^{-2}}{2!}+\cdots.$$

Then, note that $(1+z)e^{1/z} = e^{1/z}+ze^{1/z}$. Therefore,

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

Now, it should be easy to combine the sums.

• Ugh, sorry about the wonky math in the summation terms. Commented May 7, 2013 at 3:50
• Thank you very much for an explanation of the problem. By reading your post, is the Laurent series of $e^{ \frac{1}{z} }$ equal to $\sum_{-\infty}^{-1} z^{k+1}$? Commented May 7, 2013 at 3:51
• @Jamil_V Yes, but also with a $k!$ denominator that I forgot to add. I edited my post to be more correct. Commented May 7, 2013 at 3:52

The coefficients of the Laurent series is determined by

$$a_{n}=\frac{1}{2\pi i}\oint \frac{(1+z)e^{\frac{1}{z}}}{z^{n+1}}=\frac{1}{2\pi i}\oint \frac{e^{\frac{1}{z}}}{z^{n+1}}+\frac{1}{2\pi i}\oint\frac{e^{\frac{1}{z}}}{z^{n}}$$

But we know $$e^{\frac{1}{z}}=\sum_{k=0}^{\infty}\frac{1}{k!} z^{-k}$$

This means that the term $$\frac{1}{2\pi i}\cdot \frac{1}{k!}\oint\frac{e^{1/z}}{z^{n+1+k}}=\frac{1}{k!}$$ If and only if $n+1+k=-1$, which is $k=-2-n$. Similarly the second one yields $n+k=-1$, so $k=-1-n$. Since $n$ is negative this is well defined. Now we have $$a_{n}=\frac{1}{(-2-n)!}+\frac{1}{(-1-n)!}$$

• I see what is going on with the coefficients, but what do you mean by "the second one yields $n+k=-1$," if we have $n+1+k=-1$? Commented May 7, 2013 at 4:05
• @Jamil_V: Are you sure you read it in the right way? Commented May 7, 2013 at 4:06