Mixing
Determining the type of singularities
$begingroup$
Determine the type of singularities of
$$f(z)=frac{1}{(z-1)cot(pi/z)}tag{1}$$
We first rewrite the function:
$$f(z)=frac{1}{(z-1)cot(pi/z)}=frac{sin(pi/z)}{(z-1)cos(pi/z)} tag{2}$$
Now we solve
$$(z-1)cos(pi/z)overset{!}{=}0 quad Rightarrow quad hat{z}=1, z_k=frac{2}{1+2k}, quad kinmathbb Z tag{3}$$
At first, $hat{z}=1$ introduces a simple pole but then we see that $sin(pi/hat{z})=0$. So we better check the limit there:
$$lim_{zto1}frac{sin(pi/z)}{(z-1)cos(pi/z)}overset{hopital}{=}lim_{zto 1}frac{cos(pi/z)pi}{cos(pi/z)-(z-1)sin(pi/z)}=pitag{4}$$
So since that limit exists, we just found a continuation of the function at the problem point $hat{z}=1$. So we can conclude:
$hat{z}=1$ is a removable singularity.
After seeing that $sin(pi/z)$ has a root at $hat{z}=1$ we could have also just argued that the simple pole introduced by $(1-z)$ woudl have order 1 and since the first derivative of $sin(pi/z)$ doesn't have a root at $hat{z}=1$ anymore, we would have an "order" or $1-1=0$, which basically is a removable singularity.
Question: Where does this argumentation exactly come from?
Now for $z=0$ we have a non-isolated singularity, since trigonometric functions with an argument of the type $1/z$ go crazy there.
We know the Taylor series: $cos(z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n}$ so we get $$cos(1/z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n-1}=frac{1}{z}+sum_{n=1}^infty frac{(-1)^n}{(2n)!}z^{2n}tag{5}$$
So we see, for $z_k$ we get simple poles of order 1.
Anything specifically wrong here?
complex-analysis trigonometry proof-verification laurent-series singularity
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
Determine the type of singularities of
$$f(z)=frac{1}{(z-1)cot(pi/z)}tag{1}$$
We first rewrite the function:
$$f(z)=frac{1}{(z-1)cot(pi/z)}=frac{sin(pi/z)}{(z-1)cos(pi/z)} tag{2}$$
Now we solve
$$(z-1)cos(pi/z)overset{!}{=}0 quad Rightarrow quad hat{z}=1, z_k=frac{2}{1+2k}, quad kinmathbb Z tag{3}$$
At first, $hat{z}=1$ introduces a simple pole but then we see that $sin(pi/hat{z})=0$. So we better check the limit there:
$$lim_{zto1}frac{sin(pi/z)}{(z-1)cos(pi/z)}overset{hopital}{=}lim_{zto 1}frac{cos(pi/z)pi}{cos(pi/z)-(z-1)sin(pi/z)}=pitag{4}$$
So since that limit exists, we just found a continuation of the function at the problem point $hat{z}=1$. So we can conclude:
$hat{z}=1$ is a removable singularity.
After seeing that $sin(pi/z)$ has a root at $hat{z}=1$ we could have also just argued that the simple pole introduced by $(1-z)$ woudl have order 1 and since the first derivative of $sin(pi/z)$ doesn't have a root at $hat{z}=1$ anymore, we would have an "order" or $1-1=0$, which basically is a removable singularity.
Question: Where does this argumentation exactly come from?
Now for $z=0$ we have a non-isolated singularity, since trigonometric functions with an argument of the type $1/z$ go crazy there.
We know the Taylor series: $cos(z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n}$ so we get $$cos(1/z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n-1}=frac{1}{z}+sum_{n=1}^infty frac{(-1)^n}{(2n)!}z^{2n}tag{5}$$
So we see, for $z_k$ we get simple poles of order 1.
Anything specifically wrong here?
complex-analysis trigonometry proof-verification laurent-series singularity
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
$endgroup$
$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00
add a comment |
$begingroup$
Determine the type of singularities of
$$f(z)=frac{1}{(z-1)cot(pi/z)}tag{1}$$
We first rewrite the function:
$$f(z)=frac{1}{(z-1)cot(pi/z)}=frac{sin(pi/z)}{(z-1)cos(pi/z)} tag{2}$$
Now we solve
$$(z-1)cos(pi/z)overset{!}{=}0 quad Rightarrow quad hat{z}=1, z_k=frac{2}{1+2k}, quad kinmathbb Z tag{3}$$
At first, $hat{z}=1$ introduces a simple pole but then we see that $sin(pi/hat{z})=0$. So we better check the limit there:
$$lim_{zto1}frac{sin(pi/z)}{(z-1)cos(pi/z)}overset{hopital}{=}lim_{zto 1}frac{cos(pi/z)pi}{cos(pi/z)-(z-1)sin(pi/z)}=pitag{4}$$
So since that limit exists, we just found a continuation of the function at the problem point $hat{z}=1$. So we can conclude:
$hat{z}=1$ is a removable singularity.
After seeing that $sin(pi/z)$ has a root at $hat{z}=1$ we could have also just argued that the simple pole introduced by $(1-z)$ woudl have order 1 and since the first derivative of $sin(pi/z)$ doesn't have a root at $hat{z}=1$ anymore, we would have an "order" or $1-1=0$, which basically is a removable singularity.
Question: Where does this argumentation exactly come from?
Now for $z=0$ we have a non-isolated singularity, since trigonometric functions with an argument of the type $1/z$ go crazy there.
We know the Taylor series: $cos(z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n}$ so we get $$cos(1/z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n-1}=frac{1}{z}+sum_{n=1}^infty frac{(-1)^n}{(2n)!}z^{2n}tag{5}$$
So we see, for $z_k$ we get simple poles of order 1.
Anything specifically wrong here?
complex-analysis trigonometry proof-verification laurent-series singularity
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
$endgroup$
Determine the type of singularities of
$$f(z)=frac{1}{(z-1)cot(pi/z)}tag{1}$$
We first rewrite the function:
$$f(z)=frac{1}{(z-1)cot(pi/z)}=frac{sin(pi/z)}{(z-1)cos(pi/z)} tag{2}$$
Now we solve
$$(z-1)cos(pi/z)overset{!}{=}0 quad Rightarrow quad hat{z}=1, z_k=frac{2}{1+2k}, quad kinmathbb Z tag{3}$$
At first, $hat{z}=1$ introduces a simple pole but then we see that $sin(pi/hat{z})=0$. So we better check the limit there:
$$lim_{zto1}frac{sin(pi/z)}{(z-1)cos(pi/z)}overset{hopital}{=}lim_{zto 1}frac{cos(pi/z)pi}{cos(pi/z)-(z-1)sin(pi/z)}=pitag{4}$$
So since that limit exists, we just found a continuation of the function at the problem point $hat{z}=1$. So we can conclude:
$hat{z}=1$ is a removable singularity.
After seeing that $sin(pi/z)$ has a root at $hat{z}=1$ we could have also just argued that the simple pole introduced by $(1-z)$ woudl have order 1 and since the first derivative of $sin(pi/z)$ doesn't have a root at $hat{z}=1$ anymore, we would have an "order" or $1-1=0$, which basically is a removable singularity.
Question: Where does this argumentation exactly come from?
Now for $z=0$ we have a non-isolated singularity, since trigonometric functions with an argument of the type $1/z$ go crazy there.
We know the Taylor series: $cos(z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n}$ so we get $$cos(1/z)=sum_{n=0}^infty frac{(-1)^n}{(2n)!}z^{2n-1}=frac{1}{z}+sum_{n=1}^infty frac{(-1)^n}{(2n)!}z^{2n}tag{5}$$
So we see, for $z_k$ we get simple poles of order 1.
Anything specifically wrong here?
complex-analysis trigonometry proof-verification laurent-series singularity
complex-analysis trigonometry proof-verification laurent-series singularity
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
edited Feb 2 at 10:54
José Carlos Santos
174k23133243
174k23133243
asked Feb 2 at 10:43
xotixxotix
291411
asked Feb 2 at 10:43
xotixxotix
291411
asked Feb 2 at 10:43
xotixxotix
291411
291411
$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00
add a comment |
$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00
$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00
$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $sinleft(fracpi zright)$ has a simple zero at $z=1$ and $cosleft(fracpi zright)$ has not zero there,$$frac{sinleft(fracpi zright)}{(z-1)cosleft(fracpi zright)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
|
show 6 more comments
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Since $sinleft(fracpi zright)$ has a simple zero at $z=1$ and $cosleft(fracpi zright)$ has not zero there,$$frac{sinleft(fracpi zright)}{(z-1)cosleft(fracpi zright)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
|
show 6 more comments
$begingroup$
Since $sinleft(fracpi zright)$ has a simple zero at $z=1$ and $cosleft(fracpi zright)$ has not zero there,$$frac{sinleft(fracpi zright)}{(z-1)cosleft(fracpi zright)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
|
show 6 more comments
$begingroup$
Since $sinleft(fracpi zright)$ has a simple zero at $z=1$ and $cosleft(fracpi zright)$ has not zero there,$$frac{sinleft(fracpi zright)}{(z-1)cosleft(fracpi zright)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
$endgroup$
Since $sinleft(fracpi zright)$ has a simple zero at $z=1$ and $cosleft(fracpi zright)$ has not zero there,$$frac{sinleft(fracpi zright)}{(z-1)cosleft(fracpi zright)}$$has a removable singularity at $z=1$ (the simple zero in the numerator cancels the simple zero in the denominator). It's as simple as that.
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
answered Feb 2 at 10:52
José Carlos SantosJosé Carlos Santos
174k23133243
174k23133243
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
|
show 6 more comments
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
But where exactly does that come from? Is it just because of the Taylor series?
$endgroup$
– xotix
Feb 2 at 11:00
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Yes. If $1$ is a simple zero of $f$ and it is not a zero of $g$ and if the Taylor series of $f$ and $g$ centered at $1$ are $sum_{n=1}^infty a_n(z-1)^n$ and $sum_{n=0}^infty b_n(z-1)^n$ respectively, then$$frac{f(z)}{(z-1)g(z)}=frac{a_1(z-1)+a_2(z-1)^2+cdots}{b_0(z-1)+b_1(z-1)^2+cdots}=frac{a_1+a_2(z-1)+cdots}{b_0+b_1(z-1)+cdots}$$and therefore$$lim_{zto1}frac{f(z)}{g(z)}=frac{a_0}{b_1}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 11:08
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Thanks - although isn't what I wrote in (5) wrong?
$endgroup$
– xotix
Feb 2 at 20:23
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
Yes, I missed that. It should be$$cosleft(frac1zright)=sum_{n=0}^inftyfrac{(-1)^n}{(2n)!}z^{-2n}.$$
$endgroup$
– José Carlos Santos
Feb 2 at 20:46
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
$begingroup$
so that means that the Laurent series of $cos(1/z)$ has a infinite principal part, right? But since it's in the denominator, we still have a simple pole of order one?
$endgroup$
– xotix
Feb 3 at 12:33
|
show 6 more comments
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$begingroup$
Are you also an ITET student at the ETHZ in Zurich ? Bist du auch ITET student an der ETHZ ?
$endgroup$
– Poujh
Feb 3 at 22:00