Complicated Cauchy complex integral.
I have the function $f(z)$ that defined on the closed disc $ bar D_3(0)$
by the integeral on the boundry $C_3(0)$:
$f(z) = int_{C_3(0)} frac{3w^2+7w+1}{w-z}dw $
I need to find $f'(1+i)$
Caushy integral theorem and it`s conclusions tells me that on $C_3(0)$ boundry, my function is:
$f(w) = 2 pi i(3w^2+7w+1)$
and that the derivative of $f(z)$ is:
$f'(z)=int_{ C_3(0)}frac {(3w^2+7w+1)}{(w-z)^2}dw$
we use $gamma:[0,1] to Bbb C $ defined $ gamma (t)=3e^{2pi it}$ to be our path,
so we get:
$f'(1+i)=int_{0}^{1}frac {6 pi ie^{2pi i t}(3e^{4pi it}+7e^{2pi it}+1)}{(3e^{2pi it}-(1+i))^2}dt$ as integral over a path is defined.
Is there a way to pass over this complicated integral? or there is an algebric trick with whom I`ll solve this integral?, or maybe somehow to choose another parametrization to my path?
complex-analysis cauchy-integral-formula
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I have the function $f(z)$ that defined on the closed disc $ bar D_3(0)$
by the integeral on the boundry $C_3(0)$:
$f(z) = int_{C_3(0)} frac{3w^2+7w+1}{w-z}dw $
I need to find $f'(1+i)$
Caushy integral theorem and it`s conclusions tells me that on $C_3(0)$ boundry, my function is:
$f(w) = 2 pi i(3w^2+7w+1)$
and that the derivative of $f(z)$ is:
$f'(z)=int_{ C_3(0)}frac {(3w^2+7w+1)}{(w-z)^2}dw$
we use $gamma:[0,1] to Bbb C $ defined $ gamma (t)=3e^{2pi it}$ to be our path,
so we get:
$f'(1+i)=int_{0}^{1}frac {6 pi ie^{2pi i t}(3e^{4pi it}+7e^{2pi it}+1)}{(3e^{2pi it}-(1+i))^2}dt$ as integral over a path is defined.
Is there a way to pass over this complicated integral? or there is an algebric trick with whom I`ll solve this integral?, or maybe somehow to choose another parametrization to my path?
complex-analysis cauchy-integral-formula
add a comment |
I have the function $f(z)$ that defined on the closed disc $ bar D_3(0)$
by the integeral on the boundry $C_3(0)$:
$f(z) = int_{C_3(0)} frac{3w^2+7w+1}{w-z}dw $
I need to find $f'(1+i)$
Caushy integral theorem and it`s conclusions tells me that on $C_3(0)$ boundry, my function is:
$f(w) = 2 pi i(3w^2+7w+1)$
and that the derivative of $f(z)$ is:
$f'(z)=int_{ C_3(0)}frac {(3w^2+7w+1)}{(w-z)^2}dw$
we use $gamma:[0,1] to Bbb C $ defined $ gamma (t)=3e^{2pi it}$ to be our path,
so we get:
$f'(1+i)=int_{0}^{1}frac {6 pi ie^{2pi i t}(3e^{4pi it}+7e^{2pi it}+1)}{(3e^{2pi it}-(1+i))^2}dt$ as integral over a path is defined.
Is there a way to pass over this complicated integral? or there is an algebric trick with whom I`ll solve this integral?, or maybe somehow to choose another parametrization to my path?
complex-analysis cauchy-integral-formula
I have the function $f(z)$ that defined on the closed disc $ bar D_3(0)$
by the integeral on the boundry $C_3(0)$:
$f(z) = int_{C_3(0)} frac{3w^2+7w+1}{w-z}dw $
I need to find $f'(1+i)$
Caushy integral theorem and it`s conclusions tells me that on $C_3(0)$ boundry, my function is:
$f(w) = 2 pi i(3w^2+7w+1)$
and that the derivative of $f(z)$ is:
$f'(z)=int_{ C_3(0)}frac {(3w^2+7w+1)}{(w-z)^2}dw$
we use $gamma:[0,1] to Bbb C $ defined $ gamma (t)=3e^{2pi it}$ to be our path,
so we get:
$f'(1+i)=int_{0}^{1}frac {6 pi ie^{2pi i t}(3e^{4pi it}+7e^{2pi it}+1)}{(3e^{2pi it}-(1+i))^2}dt$ as integral over a path is defined.
Is there a way to pass over this complicated integral? or there is an algebric trick with whom I`ll solve this integral?, or maybe somehow to choose another parametrization to my path?
complex-analysis cauchy-integral-formula
complex-analysis cauchy-integral-formula
edited Nov 20 at 18:01
Praneet Srivastava
762516
762516
asked Nov 20 at 17:51
Daniel Vainshtein
18811
18811
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add a comment |
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You stated that $f(w) = 2pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 pi i (6w + 7)$?
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You stated that $f(w) = 2pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 pi i (6w + 7)$?
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
add a comment |
You stated that $f(w) = 2pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 pi i (6w + 7)$?
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
add a comment |
You stated that $f(w) = 2pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 pi i (6w + 7)$?
You stated that $f(w) = 2pi i (3w^2 + 7w + 1)$. Isn't it immediate that $f'(w) = 2 pi i (6w + 7)$?
answered Nov 20 at 18:07
Umberto P.
38.5k13064
38.5k13064
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
add a comment |
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
but this is only relevant on the boundry of the disc $D_3(0)$ when definitely 1+i is on the interior of $D_3(0)$
– Daniel Vainshtein
Nov 20 at 18:23
add a comment |
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