1

Consider the following functions and determine which kind of singularities they have in $z_0$. If it is a removeable singularity, then calculate the limit; if it is a pole, then give the order of the pole and the main part. $$ f(z)=\frac{1}{1-e^{z}}, z_0=0~~~~~~~~~~~~~~~~~~~~g(z)=\frac{1}{z-\sin z}, z_0=0 $$

Concerning $f(z)$, I wrote $$ f(z)=\frac{1}{\sum\limits_{i=1}^{\infty}\frac{z^i}{i!}} $$ and then considered $$ \lim\limits_{z\to 0}\frac{z^k}{\sum\limits_{i=1}^{\infty}\frac{z^i}{i!}}=0~\forall~k\in\mathbb{N}_0 $$ So the smallest $k\in\mathbb{N}_0$ for which the limit exists is $k=0$ and so here $z_0$ is a removable singularity. Is that right? And what is meant with calculating the limit now?

1 Answers1

2

I'm cringing at your use of $i$ as an index while dealing with complex numbers, but I digress.

The limit $$\lim_{z\to 0}f(z)=\lim_{z\to 0}\cfrac{-1}{\sum\limits_{n=1}^\infty\frac{z^n}{n!}}$$ fails to exist, since the denominator vanishes. (Note also the negative sign that you omitted.) However, for $z\ne 0,$ we can write $$zf(z)=\cfrac{-1}{\sum\limits_{n=1}^\infty\frac{z^{n-1}}{n!}}=\cfrac{-1}{1+\sum\limits_{m=1}^\infty\frac{z^m}{(m+1)!}},$$ whence $\lim\limits_{z\to 0}zf(z)=-1.$ Hence, $f$ has a pole of order $1$ at $z=0$. Can you take it from there?

Cameron Buie
  • 102,994
  • Is the following argumentation okay, too? Consider again $\frac{z^k}{\sum\limits_{n=1}^{\infty}\frac{z^n}{n!}}$, then this is equal to $\frac{1}{\sum\limits_{n=1}^{\infty}\frac{z^{n-k}}{n!}}$. For k=1, we have $\left\lvert\frac{z^n}{n!z}\right\rvert\leq \frac{1}{n!}$ when $z\to 0$. Because $\sum\limits_{n=1}^{\infty}\frac{1}{n!}=e-1<\infty$, we have found $k=1$ as the smallest $k\in\mathbb{N}_0$ in order to get the limit above. So we have a pole in $z_0=0$ of order 1. –  Jul 11 '13 at 13:32
  • 1
    Unfortunately, it isn't enough. This argument shows that $$\sum_{n=1}^\infty\frac{z^{n-1}}{n!}$$ converges for all $z\ne 0$ sufficiently close to $0$. However, what we must show is that $k=1$ is the least element of $\Bbb N_0$ for which $$\lim_{z\to 0}\sum_{n=1}^\infty\frac{z^{n-k}}{n!}$$ is a non-zero element of the extended complex plane (the plane together with the "point at infinity"). It is in fact the only element of $\Bbb N_0$ for which this limit is a non-zero complex number. – Cameron Buie Jul 11 '13 at 13:43
  • How can one see that k=1 is this least element? And not another k? By the way: Why does the series $\sum\limits_{m=1}^{\infty}\frac{z^m}{(m+1)!}$ converge to 0$? –  Jul 11 '13 at 13:57
  • To see that $k=1$ is the least element, show that it works for $k=1$ and doesn't for $k=0$. $\sum\limits_{m=1}^\infty\frac{z^m}{(m+1)!}$ doesn't converge to $0$, in general, but it does define a function that is continuous everywhere, and takes on a value of $0$ when $z=0$, so $$\lim_{z\to 0}\sum_{m=1}^\infty\frac{z^m}{(m+1)!}=\sum_{m=1}^\infty\frac{0^m}{(m+1)!}=0.$$ – Cameron Buie Jul 11 '13 at 14:10
  • Okay, thanks! I got it I guess. Now i know that there is a pole of order 1 in $z_0=0$ Now one has to give the "main part". Is that the main part of the Laurent-Series? I now know that for the Laurent-Series $a_{-n}=0$ for all $n>1$, but $a_{-1}\neq 0$. Fine, so $f(z)=a_{-1}z^{-1}+\sum\limits_{n=0}^{\infty}a_nz^n$. But how can I get the Laurent-Series explicitly? –  Jul 11 '13 at 14:17
  • Calculating the Residuum here? –  Jul 11 '13 at 14:28
  • I recommend finding the Taylor series of $$F(z):=\begin{cases}-1 & z=0\z\cdot f(z) & z\ne 0\end{cases}$$ about $z=0.$ Do you see how this gives us the desired Laurent series? – Cameron Buie Jul 11 '13 at 16:36
  • I only need the main part, i.e. here: $a_{-1}$. Nevertheless the Taylor series? Dont see why going this way. ;( –  Jul 11 '13 at 16:44
  • Do you mean taylorseries with development point 0? –  Jul 11 '13 at 17:20
  • Okay, I think the trick is, that $f(z)=F(z)/z$, so if one has the Taylorseries of $F(z)$ one has the Laurentseries of $f(z)$ because of the uniqueness of the Laurentseries? But my problem is: The Taylorseries of $F(z)$ with developmentpoint $a=0$ is -1 to my opinion. –  Jul 11 '13 at 17:39
  • Yes! That is precisely the connection. I don't understand your last sentence, though. – Cameron Buie Jul 11 '13 at 22:25
  • I guess my last sentence was simply stupid and false. I do not know how to determine the Taylorseries of $F(z)$... –  Jul 12 '13 at 12:41
  • It occurs to me (belatedly, of course) that I'm not sure what is meant by the "main part". Can you clarify? – Cameron Buie Jul 12 '13 at 12:55
  • Of course. If one has a Laurentseries $\sum\limits_{n=-\infty}^{\infty}a_n(z-a)^n$, then in German we call $\sum\limits_{n=1}^{\infty}a_{-n}(z-a)^{-n}$ the "Hauptteil" (english: main part) of the Laurentseries. I guess the correct name in english is "negative degree power series"? –  Jul 12 '13 at 13:01
  • Ah! Then there is no need for finding the whole Taylor series of $F(z),$ only the first term ($-1$). The Hauptteil then is $-\frac1z.$ – Cameron Buie Jul 12 '13 at 13:07
  • So $a_{-1}=-1?$ –  Jul 12 '13 at 13:18
  • Yes, indeed! Sorry to send you so far afield! – Cameron Buie Jul 12 '13 at 13:21
  • May you explain why we need only the taylorseries for the term when z=0? –  Jul 12 '13 at 13:22
  • The constant term of the Taylor series of $F$ about $z=0$--which is just $F(0)$--is the coefficient $a_{-1}$ of the Laurent series of $f$ about $z=0,$ as you've already noted. Since $f$ has a simple pole (a pole of order $1$) at $z=0$, then the residue of $f$ at $0$ is simply $$\lim_{z\to 0}zf(z)=-1.$$ I misunderstood, and thought you were looking for the whole Laurent series, so I put this in terms of the Taylor series of $F,$ instead. – Cameron Buie Jul 12 '13 at 13:29
  • As I said before, my problem is to calculate the Taylorseries of $F(z)$ because $F(z)$ has two "parts" - one, if $z=0$ and one if $z\neq 0$ - for which case do I need do develop the taylorseries ? And then it is clear that $a_{-1}$ is the constant part of this taylorseries (by comparison of the coefficients). –  Jul 12 '13 at 13:38
  • Sorry, I guess that was a stupid question! Development point is here $a=0$ because that is the critical point here... then the Taylorseries of $F(z)$ with developmentpoint $a=0$ simply is $\sum\limits_{n=0}^{\infty}\frac{F^{(n)}(0)}{n!}z^n=-1$ - right? No need to consider any cases or something like this. So, why did you wrote: The whole Taylor series? That IS the whole taylor series, isn't it? –  Jul 12 '13 at 13:47
  • So the whole Taylorseries IS -1. No more. –  Jul 12 '13 at 13:49
  • Indeed, $$F(z)=\sum_{n=0}^\infty\frac{F^{(n)}(0)}{n!}z^n$$ is the whole Taylor series for $F$ with development point $0$. Since you only want $a_{-1},$ though, we don't need anything but $\frac{F^{(0)}(0)}{0!}=F(0)=-1$. Explicitly calculating the rest of the terms $\frac{F^{(n)}(0)}{n!}$ for $n>0$ is a struggle we don't actually need to undertake. – Cameron Buie Jul 12 '13 at 13:51
  • Thank you very much for your uncomplaining help! –  Jul 12 '13 at 14:01
  • 1
    Certainly! I'm glad we managed to get past the terminology confusion! :-) – Cameron Buie Jul 12 '13 at 14:02
  • In order to finish this task: Is it right, that $g(z)$ does have an essential singularity in $z_0=0$? –  Jul 12 '13 at 17:24
  • Sorry! I revise my last question! Is it right that the function $g(z)$ has a pole of order 3 in $z_0=0$? –  Jul 12 '13 at 17:36
  • You are correct! It has a pole of order $3$. – Cameron Buie Jul 12 '13 at 22:19
  • Okay. Then I again wrote down the function $F(z):=\begin{cases}6, & z=0\z^3g(z), & z\neq 0\end{cases}$. Now the Taylorseries of $F(z)$ is $6$. So I get $g(z)=6z^{-3}$ and therefore $a_{-3}$=6. But this time I have to determine $a_{-1}, a_{-2}$, too. What's going wrong resp. how can I determine the two missing coefficients? –  Jul 13 '13 at 11:46
  • Addition: Or does that simply mean, that $a_{-1}=a_{-2}=0$? –  Jul 13 '13 at 12:07
  • Well, in this case $a_{-2}=F'(0)$ and $a_{-1}=\frac{F''(0)}{2!}.$ Same sort of situation as the last one, but we need a couple more terms of the Taylor series of $F$. – Cameron Buie Jul 13 '13 at 13:16
  • Okay, then indeed $a_{-2}=a_{-1}=0$. –  Jul 13 '13 at 14:19
  • Actually, $a_{-2}=0,$ but $a_{-1}\ne 0.$ You should find that $$F'(z)=\begin{cases}0 & z=0\ \frac{z^2(2z-3\sin z+z\cos z)}{(z-\sin z)^2} & z\ne 0,\end{cases}$$ and from this you can calculate $$F''(0)=\lim_{z\to 0}\frac{F'(z)-F(0)}{x-0}=\lim_{z\to 0}\frac{z(2z-3\sin z+z\cos z)}{(z-\sin z)^2}.$$ You should get $F''(0)=\frac35,$ so that $a_{-1}=\frac3{10}.$ – Cameron Buie Jul 13 '13 at 21:22