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Find the Laurent series expansion for $\frac{e^z}{(z+1)^2}$ for $\lvert z \rvert > 1$.

I know how to find the Laurent series expansion for $\lvert z \rvert < 1$, which is $$\frac{e^{z+1-1}}{(z+1)^2} = \frac{1}{e}\frac{e^{z+1}}{(z+1)^2}$$

and using the series expansion for $e^x$

$$\frac{1}{e} \frac{1}{(z+1)^2} \Big[ 1 + (z+1) + \frac{(z+1)^2}{2!} + ... \Big]$$

$$= \frac{1}{e} \Big[ \frac{1}{(z+1)^2} + \frac{1}{(z+1)} + \frac{1}{2!} + ... \Big]$$

So we have a pole of order $2$ and $\underset{z = -1}{Res} \frac{e^z}{(z+1)^2} = \frac{1}{e}$.

But is there any way to use this information to find the Laurent series expansion for $\lvert z \rvert > 1$? I only know how to do it if we can get a form $\frac{1}{1 - z}$ but we have an $e^z$ around so I have no idea what to do in this case.

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    $\begingroup$ The Laurent expansion about what point? In your post you have the Laurent expansion about $z=-1$, which converges in the region $\left|z+1\right|>0$. $\endgroup$ – Matt Rosenzweig May 17 '15 at 22:32
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    $\begingroup$ The Laurent expansion for $|z|<1$ should just be a Taylor series. $\endgroup$ – zhw. May 17 '15 at 22:34
  • $\begingroup$ @MattRosenzweig About $z = 0$. I guess I misunderstood how to compute a Laurent series expansion $\endgroup$ – mr eyeglasses May 17 '15 at 23:07
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The Laurent expansion about the point $c=0$ in the disk $\left|z\right|<1$ is the Cauchy product of the Taylor series of $e^{z}$ and $(z+1)^{-2}$. I think you are looking for the Laurent expansion about $c=0$ in the annulus $\left|z\right|>1$. Observe that

$$\dfrac{e^{z}}{(1+z)^{2}}=\dfrac{e^{z}}{z^{-2}(1+z^{-1})^{2}}=z^{2}e^{z}\sum_{n=0}^{\infty}(-1)^{n}z^{-n}\tag{1}$$

Since Taylor series for $e^{z}$ converges absolutely on $\mathbb{C}$ and the series $\sum (-1)^{n}z^{-n}$ converges absolutely for $\left|z\right|>1$, their product is given by

$$\sum_{m=0}^{\infty}\sum_{n=0}^{m}\dfrac{z^{n}}{n!}\dfrac{(-1)^{m-n}}{z^{m-n}}\tag{2}$$ Since the order of summation is irrelevant, we see that (2) equals

$$\begin{align*}\sum_{k=0}^{\infty}z^{k}\sum_{m=k}^{\infty}\dfrac{(-1)^{m-k}}{m!}+\sum_{k=1}^{\infty}z^{-k}\sum_{m=k}^{\infty}\dfrac{(-1)^{m}}{(m-k)!} \end{align*}\tag{3}$$

Multiplying (3) by $z^{2}$ completes the proof.

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