Laurent expansion of $frac{1}{z^2+i}$ at $z=i$
0
$begingroup$
I want to find the Laurent series of $1over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these?
$$1 over{(z+isqrt i)(z-isqrt i)}$$
$$frac{1}{i} frac{1}{1+(z^2/i)}$$
answer for this question
laurent-series
asked Dec 7 '18 at 4:23
서민규서민규
11
$endgroup$
add a comment |
0
$begingroup$
I want to find the Laurent series of $1over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these?
$$1 over{(z+isqrt i)(z-isqrt i)}$$
$$frac{1}{i} frac{1}{1+(z^2/i)}$$
answer for this question
laurent-series
asked Dec 7 '18 at 4:23
서민규서민규
11
$endgroup$
1
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51
add a comment |
0
0
0
$begingroup$
I want to find the Laurent series of $1over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these?
$$1 over{(z+isqrt i)(z-isqrt i)}$$
$$frac{1}{i} frac{1}{1+(z^2/i)}$$
answer for this question
laurent-series
asked Dec 7 '18 at 4:23
서민규서민규
11
$endgroup$
I want to find the Laurent series of $1over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these?
$$1 over{(z+isqrt i)(z-isqrt i)}$$
$$frac{1}{i} frac{1}{1+(z^2/i)}$$
answer for this question
laurent-series
laurent-series
asked Dec 7 '18 at 4:23
서민규서민규
11
asked Dec 7 '18 at 4:23
서민규서민규
11
edited Dec 7 '18 at 5:23
서민규
asked Dec 7 '18 at 4:23
서민규서민규
11
asked Dec 7 '18 at 4:23
서민규서민규
11
asked Dec 7 '18 at 4:23
서민규서민규
11
11
1
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51
add a comment |
1
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51
1
1
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51
add a comment |
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1
$begingroup$
First do partial fractions. If we abbreviate $w=(1-isqrt2)/2$, then the zeros of $z^2+i$ are $w$ and $-w$. Then you can find constants $A$ and $B$ such that $$frac1{z^2+i}=frac A{z-w}+frac B{z+w}.$$ Then you can write the denominators in the form $C+D(z-i)$ (different values of $C,D$ for the two terms), and manipulate that to a form where you can use the geometric series sum formula.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 4:55
$begingroup$
given answer is i.stack.imgur.com/ElPI7.jpg and For my solution that is laurent expansion of 1/z^2 at z=i so I think this is wrong question but im doing partial fraction for this question and i think it can be solved thx
$endgroup$
– 서민규
Dec 7 '18 at 5:38
$begingroup$
given answer is not belong to this question I think
$endgroup$
– Nosrati
Dec 7 '18 at 7:13
$begingroup$
A typo in my formula for $w$. It should read $w=(1-i)/sqrt2$ (or some other equivalent form).
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 12:51