Mixing
Calculate the Laurent Series for : $(8z+6+8i)/(2z^2-3z-4iz)$
$begingroup$
I have to find the Laurent-Series with $0<|z|<5/2$ and $z_0=0$ for:
$$frac{8z+6+8i}{2z^2-3z-4iz}$$
I already did this:
$$
frac{8z+6+8i}{2z^2-3z-4iz} = frac{A}{2cdot(z-0)}+ frac{B}{2cdot(z-1.5-2i)}\
A=-2 text{ and } B=6 \
frac{8z+6+8i}{2z^2-3z-4iz} = -1cdotfrac{1}{z} + 3cdotfrac{1}{z-1.5-2i}
$$
I hope this is correct so far.
My book says the formula for a Laurent Series is :
$$sum_{k=1}^{infty} a_kcdotfrac{1}{[z-z_0]^k} + sum_{k=0}^{infty}b_kcdot[z-z_0]^k=sum_{k=-infty}^{infty} c_{k}cdot[z-z_{0}]^{k}$$
But I have no Idea what i have to do now.
I hope someone can help me.
Thank you!
complex-analysis taylor-expansion laurent-series
edited Jan 27 at 18:13
Robert Z
101k1070143
asked Jan 27 at 17:45
kittykitty
211
$endgroup$
add a comment |
$begingroup$
I have to find the Laurent-Series with $0<|z|<5/2$ and $z_0=0$ for:
$$frac{8z+6+8i}{2z^2-3z-4iz}$$
I already did this:
$$
frac{8z+6+8i}{2z^2-3z-4iz} = frac{A}{2cdot(z-0)}+ frac{B}{2cdot(z-1.5-2i)}\
A=-2 text{ and } B=6 \
frac{8z+6+8i}{2z^2-3z-4iz} = -1cdotfrac{1}{z} + 3cdotfrac{1}{z-1.5-2i}
$$
I hope this is correct so far.
My book says the formula for a Laurent Series is :
$$sum_{k=1}^{infty} a_kcdotfrac{1}{[z-z_0]^k} + sum_{k=0}^{infty}b_kcdot[z-z_0]^k=sum_{k=-infty}^{infty} c_{k}cdot[z-z_{0}]^{k}$$
But I have no Idea what i have to do now.
I hope someone can help me.
Thank you!
complex-analysis taylor-expansion laurent-series
edited Jan 27 at 18:13
Robert Z
101k1070143
asked Jan 27 at 17:45
kittykitty
211
$endgroup$
add a comment |
$begingroup$
I have to find the Laurent-Series with $0<|z|<5/2$ and $z_0=0$ for:
$$frac{8z+6+8i}{2z^2-3z-4iz}$$
I already did this:
$$
frac{8z+6+8i}{2z^2-3z-4iz} = frac{A}{2cdot(z-0)}+ frac{B}{2cdot(z-1.5-2i)}\
A=-2 text{ and } B=6 \
frac{8z+6+8i}{2z^2-3z-4iz} = -1cdotfrac{1}{z} + 3cdotfrac{1}{z-1.5-2i}
$$
I hope this is correct so far.
My book says the formula for a Laurent Series is :
$$sum_{k=1}^{infty} a_kcdotfrac{1}{[z-z_0]^k} + sum_{k=0}^{infty}b_kcdot[z-z_0]^k=sum_{k=-infty}^{infty} c_{k}cdot[z-z_{0}]^{k}$$
But I have no Idea what i have to do now.
I hope someone can help me.
Thank you!
complex-analysis taylor-expansion laurent-series
edited Jan 27 at 18:13
Robert Z
101k1070143
asked Jan 27 at 17:45
kittykitty
211
$endgroup$
I have to find the Laurent-Series with $0<|z|<5/2$ and $z_0=0$ for:
$$frac{8z+6+8i}{2z^2-3z-4iz}$$
I already did this:
$$
frac{8z+6+8i}{2z^2-3z-4iz} = frac{A}{2cdot(z-0)}+ frac{B}{2cdot(z-1.5-2i)}\
A=-2 text{ and } B=6 \
frac{8z+6+8i}{2z^2-3z-4iz} = -1cdotfrac{1}{z} + 3cdotfrac{1}{z-1.5-2i}
$$
I hope this is correct so far.
My book says the formula for a Laurent Series is :
$$sum_{k=1}^{infty} a_kcdotfrac{1}{[z-z_0]^k} + sum_{k=0}^{infty}b_kcdot[z-z_0]^k=sum_{k=-infty}^{infty} c_{k}cdot[z-z_{0}]^{k}$$
But I have no Idea what i have to do now.
I hope someone can help me.
Thank you!
complex-analysis taylor-expansion laurent-series
complex-analysis taylor-expansion laurent-series
edited Jan 27 at 18:13
Robert Z
101k1070143
asked Jan 27 at 17:45
kittykitty
211
edited Jan 27 at 18:13
Robert Z
101k1070143
asked Jan 27 at 17:45
kittykitty
211
edited Jan 27 at 18:13
Robert Z
101k1070143
edited Jan 27 at 18:13
Robert Z
101k1070143
edited Jan 27 at 18:13
Robert Z
101k1070143
101k1070143
asked Jan 27 at 17:45
kittykitty
211
asked Jan 27 at 17:45
kittykitty
211
asked Jan 27 at 17:45
kittykitty
211
211
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
You are on the right track. After the decomposition (please check your computations),
$$frac{8z+6+8i}{2z^2-3z-4iz}=-frac{2}{z}+frac{12}{2z-(3+4i)}=
-frac{2}{z}-frac{frac{12}{3+4i}}{1-frac{2z}{(3+4i)}}$$
you should expand $(1-frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=sqrt{2^2+4^2}/2=5/2$) as
$$sum_{k=0}^{infty}left(frac{2z}{(3+4i)}right)^k.$$
Can you take it from here?
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
You are on the right track. After the decomposition (please check your computations),
$$frac{8z+6+8i}{2z^2-3z-4iz}=-frac{2}{z}+frac{12}{2z-(3+4i)}=
-frac{2}{z}-frac{frac{12}{3+4i}}{1-frac{2z}{(3+4i)}}$$
you should expand $(1-frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=sqrt{2^2+4^2}/2=5/2$) as
$$sum_{k=0}^{infty}left(frac{2z}{(3+4i)}right)^k.$$
Can you take it from here?
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
add a comment |
$begingroup$
You are on the right track. After the decomposition (please check your computations),
$$frac{8z+6+8i}{2z^2-3z-4iz}=-frac{2}{z}+frac{12}{2z-(3+4i)}=
-frac{2}{z}-frac{frac{12}{3+4i}}{1-frac{2z}{(3+4i)}}$$
you should expand $(1-frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=sqrt{2^2+4^2}/2=5/2$) as
$$sum_{k=0}^{infty}left(frac{2z}{(3+4i)}right)^k.$$
Can you take it from here?
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
add a comment |
$begingroup$
You are on the right track. After the decomposition (please check your computations),
$$frac{8z+6+8i}{2z^2-3z-4iz}=-frac{2}{z}+frac{12}{2z-(3+4i)}=
-frac{2}{z}-frac{frac{12}{3+4i}}{1-frac{2z}{(3+4i)}}$$
you should expand $(1-frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=sqrt{2^2+4^2}/2=5/2$) as
$$sum_{k=0}^{infty}left(frac{2z}{(3+4i)}right)^k.$$
Can you take it from here?
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
$endgroup$
You are on the right track. After the decomposition (please check your computations),
$$frac{8z+6+8i}{2z^2-3z-4iz}=-frac{2}{z}+frac{12}{2z-(3+4i)}=
-frac{2}{z}-frac{frac{12}{3+4i}}{1-frac{2z}{(3+4i)}}$$
you should expand $(1-frac{2z}{(3+4i)})^{-1}$ at $z=0$ (note that it is holomorphic in the disc $|z|< |(3+4i)/2|=sqrt{2^2+4^2}/2=5/2$) as
$$sum_{k=0}^{infty}left(frac{2z}{(3+4i)}right)^k.$$
Can you take it from here?
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
edited Jan 28 at 11:04
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
answered Jan 27 at 18:02
Robert ZRobert Z
101k1070143
101k1070143
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
add a comment |
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
Thanks for your fast answer. I am completly confused now. I thought I would take the two parts (the A and he B part) and bring them into the form of of the formular? But why ignore $-frac{2}{z}$ and only go with the B part? Why change the second part? I thought 1/(z-something) would be the right approach to get it into the formular. Sorry I really don't understand whats happening here.
$endgroup$
– kitty
Jan 27 at 18:54
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
$begingroup$
No, you have to consider also $2/z$ which is ready for the Laurent expansion. Note that here $z_0=0$ so the Laurent expansion has the form $sum_{k=1}^{infty}frac{a_k}{z^k} + sum_{k=0}^{infty}b_kcdot z^k$.
$endgroup$
– Robert Z
Jan 27 at 19:06
add a comment |
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