# Find the principal part at poles

I am asked to find the isolated singularities of the function and determine their typing, order, and finding the principal part at each pole. So for the first function, which is $$\frac{e^z-e}{z^2-1}$$, I know that it has poles of order $$1$$ at $$\pm1$$. However, when trying to find the principal part, I'm running into some issues.

My attempt was to rewrite as a partial, $$\frac{e^z-e}{z^2-1} = -\frac{e^z-e}{2(z+1)} + \frac{e^z-e}{2(z-1)}$$

But then I'm not really sure where to go from there. I suppose you're supposed to try and convert it to a Laurent series, but I'm still kind of confused on how to go about that. Any hints or help is appreciated.

Edit:

Tried to keep pushing on. Here's what I got:

$$-\frac{e^z-e}{2(z+1)} + \frac{e^z-e}{2(z-1)} = -\frac{e^z-e}{2}(\frac{1}{1-(-z)}) - \frac{e^z-e}{2}(\frac{1}{1-z})$$

$$\Rightarrow -\frac{e^z-e}{2} ( \sum_{k=0}^{\infty}(-z)^k + \sum_{k=0}^{\infty} z^k )$$

But I have a feeling I've done something wrong, as it appears that no principal part pops out (all exponential values will be positive). Did I do something wrong?

For each $$z\in\Bbb C$$, you have\begin{align}e^z-e&=e\left(e^{z-1}-1\right)\\&=e\left((z-1)+\frac1{2!}(z-1)^2+\frac1{3!}(z-1)^3+\cdots\right),\end{align}and therefore, if $$z\ne1$$,$$\frac{e^z-e}{z-1}=e(z-1)+\frac e{2!}(z-1)^2+\frac e{3!}(z-1)^3+\cdots$$On the other hand, if $$|z-1|<2$$\begin{align}\frac1{z+1}&=\frac1{2+(z-1)}\\&=\frac12\cdot\frac1{1+(z-1)/2}\\&=\frac12\sum_{n=0}^\infty(-1)^n\left(\frac{z-1}2\right)^n\\&=\sum_{n=0}^\infty(-1)^n\frac{(z-1)^n}{2^{n+1}},\end{align}and therefore\begin{align}\frac{e^z-e}{z+1}&=\left(e(z-1)+\frac e{2!}(z-1)^2+\frac e{3!}(z-1)^3+\cdots\right)\left(\frac12-\frac{z-1}{2^2}+\frac{(z-1)^2}{2^3}-\cdots\right)\\&=\frac e2+\frac e{12}(z-1)^3-\frac e{48}(z-1)^4+\cdots\end{align}

At $$-1$$,$$\dfrac{e^z-e}{z-1}=\frac12\left(e-\frac1e\right).$$Therefore, the principal part of your function at $$-1$$ is$$\frac12\left(e-\frac1e\right)\frac1{z+1},$$and you have a pole of order $$1$$ there.

• And so there is no principal part? Apr 17 at 22:36
• Indeed, in both cases. Apr 17 at 22:37
• Gotcha. And how did you arrive at the domain restriction $\vert z-1 \vert \lt 2$? And furthermore how did you simplify the original function to $\frac{e^z-e}{z-1}$? What happened to the $z+1$ in the denominator? Sorry, I guess I'm really confused about how you simplified the function in general. Apr 17 at 22:57
• I forgot to deal with the case $z=-1$; I've added that. And I arrived at the restriction $|z-1|<2$ in order to have the equality$$\frac1{1+(z-1)/2}=\sum_{n=0}^\infty(-1)^n\left(\frac{z-1}2\right)^n.$$ Apr 18 at 6:07

$$\frac{1}{z+1} = \frac{1}{((z-1)+2)} = \frac{1}{2\left(\frac{(z-1)}{2}+1\right)} = \frac{1}{2}\sum_{j=0}^{\infty}\frac{(-1)^j(z-1)^j}{2^j} \quad |z-1|<2$$

Hence $$\frac{1}{(z-1)(z+1)} = \frac{1}{2}\sum_{j=0}^{\infty}\frac{(-1)^j(z-1)^{j-1}}{2^j} \quad |z-1|<2 \tag{1}$$

Using the same trick as José:

$$e(e^{z-1}-1) = e\sum_{k=0}^{\infty} \frac{(z-1)^{k+1}}{(k+1)!} \tag{2}$$

By the Cauchy product of (1) and (2):

$$\frac{e^{z}-e}{z^2-1} = e\sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{(-1)^{n-k}(z-1)^{n}}{(k+1)!2^{n-k+1}} = e\sum_{n=0}^{\infty}\left[ \left(\sum_{k=0}^{n}\frac{(-1)^{n-k}}{(k+1)!2^{n-k+1}} \right)(z-1)^{n}\right] \quad |z-1|<2$$

Since $$n\geq0$$ we have no principal part at $$z=1$$

For $$z=-1$$ use:

$$\frac{1}{z-1}= \frac{1}{(z+1)-2} = \frac{1}{2\left(\frac{(z+1)}{2}-1\right)} = -\frac{1}{2}\sum_{j=0}^{\infty}\frac{(z+1)^j}{2^j} \quad |z+1|<2$$

Hence

$$\frac{1}{(z-1)(z+1)}= -\frac{1}{2}\sum_{j=0}^{\infty}\frac{(z+1)^{j-1}}{2^j} \quad |z+1|<2 \tag{3}$$

$$e^{-1}(e^{z+1}-1+(1-e^2)) = e^{-1}\sum_{k=0}^{\infty} \frac{(z+1)^{k+1}}{(k+1)!} + (e^{-1}-e) \tag{4}$$

By the Cauchy product of (3) and (4):

$$\frac{e^{z}-e}{z^2-1} = -e^{-1}\sum_{n=0}^{\infty}\left[ \left(\sum_{k=0}^{n}\frac{1}{(k+1)!2^{n-k+1}} \right)(z+1)^{n}\right]- \frac{e^{-1}-e}{2}\sum_{j=0}^{\infty}\frac{(z+1)^{j-1}}{2^j} \quad |z+1|<2$$

Note that for $$j<1$$ ie $$j=0$$ we have a principal part: $$\boxed{\displaystyle \frac{e-e^{-1}}{2(z+1)}}$$

• Where does your answer differ from José's, saying that there exists a principal part? Apr 18 at 0:16
• Hi @robertlewison . There is a principal part for $z=−1$. Apart from that, the Cauchy product helps to visualize the Laurent expansion in a compact formula. Cheers. Apr 18 at 0:21

Notice that $$\lim_{z \to 1} \frac{e^z - e}{z^2 - 1} = \lim_{z \to 1} \frac{\frac{e^z - e}{z - 1}}{\frac{z^2 - 1}{z - 1}} = \frac{\lim_{z \to 1} \frac{e^z - e}{z - 1}}{\lim_{z \to 1} \frac{z^2 - 1}{z - 1}} = \frac{\left.\frac{d}{dz}\right\vert_{z = 1} \exp z}{\left.\frac{d}{dz}\right\vert_{z = 1} (z^2)} = \frac{e}{2} ,$$ so the singularity at $$z = 1$$ is in fact removable.

For the pole at $$z = -1$$, write $$w := z + 1$$, so that the function is $$-\frac{e^{w - 1} - e}{2 w} + g(w)$$ for some function analytic near $$w$$ (which hence doesn't affect the principal part). If we expand the numerator of the first term in a Taylor series about $$w = 0$$, only the constant term can contribute to the principal part (why?).

Expanding gives $$e^{w - 1} - e = e^{0 - 1} - e + R(w) = \frac{1}{e} - e + R(w)$$, where $$R(w) \in O(w)$$, so the principal part of the Laurent series of the function on a punctured disk at $$w = 0$$ is $$\frac{e - \frac1e}{2 w} = \boxed{\frac{\sinh 1}{z - 1}} .$$

• I see. Could you explain your step where you went to $$\frac{\left.\frac{d}{dz}\right\vert_{z = 1} \exp z}{\left.\frac{d}{dz}\right\vert_{z = 1} (z^2 - 1)}$$ I think I might be confused about your notation. Apr 17 at 22:29
• @robertlewison You mean how to get that expression? I think he used the definition of limit - for eg, $\left.\frac{d}{dz}\right\vert_{z = 1} (z^2 - 1)$ is defined to be $\lim_{z \to 1} \frac{z^2 - 1}{z - 1}$.
– Tri
Apr 17 at 22:47
• I'm just kind of confused as to where the derivative is coming from. Have a feeling I might be missing a pretty foundational step. Apr 17 at 22:51
• @robertlewison as I said above, the derivatives appear because the numerator and denominators are just definitions of calculating derivative of functions at $z=1$.
– Tri
Apr 17 at 22:59
• Right, to show that the singularity at $1$ is removable, it's enough to show that the limit of the quantity as $z \to 1$ is finite. The key observation is that both the numerator and the denominator of the original expression, $\frac{e^z - e}{z^2 - 1}$ look like the numerators of difference quotients of familiar functions. But the limit of a difference quotient is, by definition, a derivative. Apr 18 at 0:30