Finding the Laurent Series around a given point
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While studying I got this exercise:
Find the Laurent Series expansion valid for $0 < |z - i| < sqrt2$ for the following function:$$f(z) = frac{1}{(z-i)^8(z+1)}$$
So I have to get a series expansion around the point $i$.
I tried using $w = z - i$ to see if I could get somewhere but all that I got was $$f(w) = frac{1}{w^8}*frac{1}{w + i + 1}$$ which doesn't help me much because I can't turn the second fraction into a geometric series, I thought that maybe I was supposed to use the $|1 + i|$ especially because that evaluates to $sqrt2$ but I don't really know if there is a valid way to do that.
Appreciate any help.
Edit: From AndreasBlass comment I got it to $$frac{1}{i+1}sum_{n=0}^infty frac{w^{n-8}}{(1+i)^n}$$ which I think would indeed be valid for the range they want, because as I said $|1 + i| = sqrt2$
complex-analysis power-series laurent-series
asked Jan 15 at 20:38
João DavidJoão David
255
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add a comment |
$begingroup$
While studying I got this exercise:
Find the Laurent Series expansion valid for $0 < |z - i| < sqrt2$ for the following function:$$f(z) = frac{1}{(z-i)^8(z+1)}$$
So I have to get a series expansion around the point $i$.
I tried using $w = z - i$ to see if I could get somewhere but all that I got was $$f(w) = frac{1}{w^8}*frac{1}{w + i + 1}$$ which doesn't help me much because I can't turn the second fraction into a geometric series, I thought that maybe I was supposed to use the $|1 + i|$ especially because that evaluates to $sqrt2$ but I don't really know if there is a valid way to do that.
Appreciate any help.
Edit: From AndreasBlass comment I got it to $$frac{1}{i+1}sum_{n=0}^infty frac{w^{n-8}}{(1+i)^n}$$ which I think would indeed be valid for the range they want, because as I said $|1 + i| = sqrt2$
complex-analysis power-series laurent-series
asked Jan 15 at 20:38
João DavidJoão David
255
$endgroup$
2
$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
1
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45
add a comment |
$begingroup$
While studying I got this exercise:
Find the Laurent Series expansion valid for $0 < |z - i| < sqrt2$ for the following function:$$f(z) = frac{1}{(z-i)^8(z+1)}$$
So I have to get a series expansion around the point $i$.
I tried using $w = z - i$ to see if I could get somewhere but all that I got was $$f(w) = frac{1}{w^8}*frac{1}{w + i + 1}$$ which doesn't help me much because I can't turn the second fraction into a geometric series, I thought that maybe I was supposed to use the $|1 + i|$ especially because that evaluates to $sqrt2$ but I don't really know if there is a valid way to do that.
Appreciate any help.
Edit: From AndreasBlass comment I got it to $$frac{1}{i+1}sum_{n=0}^infty frac{w^{n-8}}{(1+i)^n}$$ which I think would indeed be valid for the range they want, because as I said $|1 + i| = sqrt2$
complex-analysis power-series laurent-series
asked Jan 15 at 20:38
João DavidJoão David
255
$endgroup$
While studying I got this exercise:
Find the Laurent Series expansion valid for $0 < |z - i| < sqrt2$ for the following function:$$f(z) = frac{1}{(z-i)^8(z+1)}$$
So I have to get a series expansion around the point $i$.
I tried using $w = z - i$ to see if I could get somewhere but all that I got was $$f(w) = frac{1}{w^8}*frac{1}{w + i + 1}$$ which doesn't help me much because I can't turn the second fraction into a geometric series, I thought that maybe I was supposed to use the $|1 + i|$ especially because that evaluates to $sqrt2$ but I don't really know if there is a valid way to do that.
Appreciate any help.
Edit: From AndreasBlass comment I got it to $$frac{1}{i+1}sum_{n=0}^infty frac{w^{n-8}}{(1+i)^n}$$ which I think would indeed be valid for the range they want, because as I said $|1 + i| = sqrt2$
complex-analysis power-series laurent-series
complex-analysis power-series laurent-series
asked Jan 15 at 20:38
João DavidJoão David
255
asked Jan 15 at 20:38
João DavidJoão David
255
edited Jan 15 at 20:56
João David
asked Jan 15 at 20:38
João DavidJoão David
255
asked Jan 15 at 20:38
João DavidJoão David
255
asked Jan 15 at 20:38
João DavidJoão David
255
255
2
$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
1
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45
add a comment |
2
$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
1
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45
2
2
$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
1
1
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45
add a comment |
1 Answer
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As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$frac{1}{w^8(1+i)}frac{1}{1 + frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.
answered Jan 15 at 20:54
João DavidJoão David
255
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add a comment |
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$begingroup$
As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$frac{1}{w^8(1+i)}frac{1}{1 + frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.
answered Jan 15 at 20:54
João DavidJoão David
255
$endgroup$
add a comment |
$begingroup$
As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$frac{1}{w^8(1+i)}frac{1}{1 + frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.
answered Jan 15 at 20:54
João DavidJoão David
255
$endgroup$
add a comment |
$begingroup$
As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$frac{1}{w^8(1+i)}frac{1}{1 + frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.
answered Jan 15 at 20:54
João DavidJoão David
255
$endgroup$
As I've updated the main post I just want to explain my confusion, which came from mostly seeing problems where I had fractions that looked like $$frac{1}{a - r}$$ where a was simply a real number.
The solution came from, as AndreasBlass pointed, using a complex number instead of the real number, which putting it in evidence gave me $$frac{1}{w^8(1+i)}frac{1}{1 + frac{w}{1+i}}$$ which would converge if $|w| < |1 + i|$.
answered Jan 15 at 20:54
João DavidJoão David
255
answered Jan 15 at 20:54
João DavidJoão David
255
answered Jan 15 at 20:54
João DavidJoão David
255
answered Jan 15 at 20:54
João DavidJoão David
255
255
add a comment |
add a comment |
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$begingroup$
What exactly prevented you from turning the second fraction into a geometric series? Have you previously turned fractions of the form $1/(z-r)$ into geometric series? If so, how is the present example different from ones where you succeeded?
$endgroup$
– Andreas Blass
Jan 15 at 20:43
1
$begingroup$
@AndreasBlass Oh I think I see where it is getting, will update in a bit
$endgroup$
– João David
Jan 15 at 20:45