Laurent Expansion Of $frac1z^2+1$

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Expand $frac1z^2+1$ around $0<|z-i|<2$

$$frac1z^2+1=frac12i(z-i)-frac12i(z+i)=frac12icdotfrac1(z-i)-frac12icdotfrac1(z+i)$$

How can I expand it to Laurent series?

complex-analysis laurent-series

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edited Jul 16 at 11:55

Michael Hardy

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asked Jul 16 at 11:49

newhere

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up vote
0
down vote

favorite

Expand $frac1z^2+1$ around $0<|z-i|<2$

$$frac1z^2+1=frac12i(z-i)-frac12i(z+i)=frac12icdotfrac1(z-i)-frac12icdotfrac1(z+i)$$

How can I expand it to Laurent series?

complex-analysis laurent-series

share|cite|improve this question

edited Jul 16 at 11:55

Michael Hardy

204k23186463

asked Jul 16 at 11:49

newhere

759310

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Expand $frac1z^2+1$ around $0<|z-i|<2$

$$frac1z^2+1=frac12i(z-i)-frac12i(z+i)=frac12icdotfrac1(z-i)-frac12icdotfrac1(z+i)$$

How can I expand it to Laurent series?

complex-analysis laurent-series

share|cite|improve this question

edited Jul 16 at 11:55

Michael Hardy

204k23186463

asked Jul 16 at 11:49

newhere

759310

Expand $frac1z^2+1$ around $0<|z-i|<2$

$$frac1z^2+1=frac12i(z-i)-frac12i(z+i)=frac12icdotfrac1(z-i)-frac12icdotfrac1(z+i)$$

How can I expand it to Laurent series?

complex-analysis laurent-series

share|cite|improve this question

edited Jul 16 at 11:55

Michael Hardy

204k23186463

asked Jul 16 at 11:49

newhere

759310

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 11:55

Michael Hardy

204k23186463

edited Jul 16 at 11:55

Michael Hardy

204k23186463

edited Jul 16 at 11:55

Michael Hardy

204k23186463

204k23186463

asked Jul 16 at 11:49

newhere

759310

asked Jul 16 at 11:49

newhere

759310

asked Jul 16 at 11:49

newhere

759310

759310

add a comment | 

add a comment | 

1 Answer
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We have the complex function with respect to $z$ :

$$f(z) = frac1z^2+1$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = frac1z^2+1 = frac1(z+i)(z-i)=frac1[(z-i)+2i](z-i)$$
$$=$$
$$frac1(z-i)^2big[1+frac2iz-ibig]$$

Recall now, that :

$$frac11+w = sum_n=0^infty (-1)^nw^n, ; ; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = frac1(z-i)^2cdot frac11+frac2iz-i=frac1(z-i)^2sum_n=0^infty(-1)^nbigg(frac2iz-ibigg)^n$$
$$implies$$
$$f(z) = sum_n=0^infty(-1)^nfrac(2i)^n(z-i)^n+2$$

for $bigg| frac2iz-i bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.

share|cite|improve this answer

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
  – Riemann
  Jul 16 at 13:01
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote

We have the complex function with respect to $z$ :

$$f(z) = frac1z^2+1$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = frac1z^2+1 = frac1(z+i)(z-i)=frac1[(z-i)+2i](z-i)$$
$$=$$
$$frac1(z-i)^2big[1+frac2iz-ibig]$$

Recall now, that :

$$frac11+w = sum_n=0^infty (-1)^nw^n, ; ; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = frac1(z-i)^2cdot frac11+frac2iz-i=frac1(z-i)^2sum_n=0^infty(-1)^nbigg(frac2iz-ibigg)^n$$
$$implies$$
$$f(z) = sum_n=0^infty(-1)^nfrac(2i)^n(z-i)^n+2$$

for $bigg| frac2iz-i bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.

share|cite|improve this answer

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
  – Riemann
  Jul 16 at 13:01
  

  
  
  
  

add a comment | 

up vote
5
down vote

We have the complex function with respect to $z$ :

$$f(z) = frac1z^2+1$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = frac1z^2+1 = frac1(z+i)(z-i)=frac1[(z-i)+2i](z-i)$$
$$=$$
$$frac1(z-i)^2big[1+frac2iz-ibig]$$

Recall now, that :

$$frac11+w = sum_n=0^infty (-1)^nw^n, ; ; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = frac1(z-i)^2cdot frac11+frac2iz-i=frac1(z-i)^2sum_n=0^infty(-1)^nbigg(frac2iz-ibigg)^n$$
$$implies$$
$$f(z) = sum_n=0^infty(-1)^nfrac(2i)^n(z-i)^n+2$$

for $bigg| frac2iz-i bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.

share|cite|improve this answer

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
  – Riemann
  Jul 16 at 13:01
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

We have the complex function with respect to $z$ :

$$f(z) = frac1z^2+1$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = frac1z^2+1 = frac1(z+i)(z-i)=frac1[(z-i)+2i](z-i)$$
$$=$$
$$frac1(z-i)^2big[1+frac2iz-ibig]$$

Recall now, that :

$$frac11+w = sum_n=0^infty (-1)^nw^n, ; ; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = frac1(z-i)^2cdot frac11+frac2iz-i=frac1(z-i)^2sum_n=0^infty(-1)^nbigg(frac2iz-ibigg)^n$$
$$implies$$
$$f(z) = sum_n=0^infty(-1)^nfrac(2i)^n(z-i)^n+2$$

for $bigg| frac2iz-i bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.

share|cite|improve this answer

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

We have the complex function with respect to $z$ :

$$f(z) = frac1z^2+1$$

Note, that to form the factor $z-i$, one can manipulate $f(z)$, as :

$$f(z) = frac1z^2+1 = frac1(z+i)(z-i)=frac1[(z-i)+2i](z-i)$$
$$=$$
$$frac1(z-i)^2big[1+frac2iz-ibig]$$

Recall now, that :

$$frac11+w = sum_n=0^infty (-1)^nw^n, ; ; |w|<1$$

thus, the function can be written as an expansion :

$$f(z) = frac1(z-i)^2cdot frac11+frac2iz-i=frac1(z-i)^2sum_n=0^infty(-1)^nbigg(frac2iz-ibigg)^n$$
$$implies$$
$$f(z) = sum_n=0^infty(-1)^nfrac(2i)^n(z-i)^n+2$$

for $bigg| frac2iz-i bigg| < 1 $.

Since the expansion of an analytic complex function is unique, this is the Laurent Expansion wanted.

share|cite|improve this answer

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 13:11

edited Jul 16 at 13:11

edited Jul 16 at 13:11

answered Jul 16 at 12:02

Rebellos

10k21039

answered Jul 16 at 12:02

Rebellos

10k21039

answered Jul 16 at 12:02

Rebellos

10k21039

10k21039

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
  – Riemann
  Jul 16 at 13:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
  – Riemann
  Jul 16 at 13:01
  

  
  
  
  

1

1

Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
– Riemann
Jul 16 at 13:01

Are you sure $bigg| frac2iz-i bigg| < 1 Leftrightarrow 0 < |z-i| < 2$ ??
– Riemann
Jul 16 at 13:01

add a comment | 

 

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