Mixing
Expand complex function into series (factoring not working)
$begingroup$
$frac{z}{z^2 + 9}$, I need to turn into series, where z = complex number.
I have tried factoring denominator to get $frac{1}{z} cdot frac{1}{(1 + frac{9}{z^2})}$ and use binomial expansion. But solution gives answer in powers of $(frac{z}{9})$. What am I doing wrong?
complex-analysis laurent-series
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
$endgroup$
add a comment |
$begingroup$
$frac{z}{z^2 + 9}$, I need to turn into series, where z = complex number.
I have tried factoring denominator to get $frac{1}{z} cdot frac{1}{(1 + frac{9}{z^2})}$ and use binomial expansion. But solution gives answer in powers of $(frac{z}{9})$. What am I doing wrong?
complex-analysis laurent-series
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
$endgroup$
add a comment |
$begingroup$
$frac{z}{z^2 + 9}$, I need to turn into series, where z = complex number.
I have tried factoring denominator to get $frac{1}{z} cdot frac{1}{(1 + frac{9}{z^2})}$ and use binomial expansion. But solution gives answer in powers of $(frac{z}{9})$. What am I doing wrong?
complex-analysis laurent-series
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
$endgroup$
$frac{z}{z^2 + 9}$, I need to turn into series, where z = complex number.
I have tried factoring denominator to get $frac{1}{z} cdot frac{1}{(1 + frac{9}{z^2})}$ and use binomial expansion. But solution gives answer in powers of $(frac{z}{9})$. What am I doing wrong?
complex-analysis laurent-series
complex-analysis laurent-series
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
edited Jan 20 at 3:10
MinYoung Kim
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
asked Jan 20 at 3:00
MinYoung KimMinYoung Kim
907
907
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
What about partial fraction?
$$frac{z}{z^2 + 9} = frac {1}{2} (frac {1}{z+3i} + frac {1}{z-3i})$$
$$frac {1}{z+3i} = frac {1}{3i} (frac {1}{1+(frac {z}{3i})})$$
Use geometric series for the above and similarly for the other part.
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
add a comment |
$begingroup$
$$frac{z}{z^2 + 9} = z (frac{1}{z^2 + 9}) = frac {z}{9} ( frac {1}{1+(frac {z^2}{9})})$$
$$ frac {1}{1+(frac {z^2}{9})} = sum _0 ^ infty (-1)^n (frac {z^2}{9})^n $$
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
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$begingroup$
What about partial fraction?
$$frac{z}{z^2 + 9} = frac {1}{2} (frac {1}{z+3i} + frac {1}{z-3i})$$
$$frac {1}{z+3i} = frac {1}{3i} (frac {1}{1+(frac {z}{3i})})$$
Use geometric series for the above and similarly for the other part.
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
add a comment |
$begingroup$
What about partial fraction?
$$frac{z}{z^2 + 9} = frac {1}{2} (frac {1}{z+3i} + frac {1}{z-3i})$$
$$frac {1}{z+3i} = frac {1}{3i} (frac {1}{1+(frac {z}{3i})})$$
Use geometric series for the above and similarly for the other part.
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
add a comment |
$begingroup$
What about partial fraction?
$$frac{z}{z^2 + 9} = frac {1}{2} (frac {1}{z+3i} + frac {1}{z-3i})$$
$$frac {1}{z+3i} = frac {1}{3i} (frac {1}{1+(frac {z}{3i})})$$
Use geometric series for the above and similarly for the other part.
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
What about partial fraction?
$$frac{z}{z^2 + 9} = frac {1}{2} (frac {1}{z+3i} + frac {1}{z-3i})$$
$$frac {1}{z+3i} = frac {1}{3i} (frac {1}{1+(frac {z}{3i})})$$
Use geometric series for the above and similarly for the other part.
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:13
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
add a comment |
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
$begingroup$
Thank you. I am trying to understand why the book gave answers in terms of powers of (z/9), what was wrong with my method? One guess is that the area of convergence changes from inside the "circle" in complex plane to outside the "circle" when I make the series as powers of $frac{9}{z^2}$, although the series was expanded from the same original equation..
$endgroup$
– MinYoung Kim
Jan 20 at 3:34
add a comment |
$begingroup$
$$frac{z}{z^2 + 9} = z (frac{1}{z^2 + 9}) = frac {z}{9} ( frac {1}{1+(frac {z^2}{9})})$$
$$ frac {1}{1+(frac {z^2}{9})} = sum _0 ^ infty (-1)^n (frac {z^2}{9})^n $$
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
$$frac{z}{z^2 + 9} = z (frac{1}{z^2 + 9}) = frac {z}{9} ( frac {1}{1+(frac {z^2}{9})})$$
$$ frac {1}{1+(frac {z^2}{9})} = sum _0 ^ infty (-1)^n (frac {z^2}{9})^n $$
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
$$frac{z}{z^2 + 9} = z (frac{1}{z^2 + 9}) = frac {z}{9} ( frac {1}{1+(frac {z^2}{9})})$$
$$ frac {1}{1+(frac {z^2}{9})} = sum _0 ^ infty (-1)^n (frac {z^2}{9})^n $$
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$$frac{z}{z^2 + 9} = z (frac{1}{z^2 + 9}) = frac {z}{9} ( frac {1}{1+(frac {z^2}{9})})$$
$$ frac {1}{1+(frac {z^2}{9})} = sum _0 ^ infty (-1)^n (frac {z^2}{9})^n $$
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 20 at 3:23
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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