Is it possible to prove Cauchy integral formula by parametrizing the circle?
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Theorem. Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0) = frac{1}{2pi i} int_C frac{f(z)dz}{z-z_0}.$$
The proof starts with letting $C_r$ denote a positively oriented circle $|z-z_0|=r$, where $r$ is small enough that $C_r$ is interior to $C$. Since the function $f(z)/(z-z_0)$ is analytic between and on the contours $C$ and $C_r$, we have $$ int_C frac{f(z)dz}{z-z_0} = int_{C_r} frac{f(z)dz}{z-z_0}.$$
Can we then parametrize the contour $C_r$ and calculate $int_{C_r} frac{f(z)dz}{z-z_0}$? How to proceed?
calculus complex-numbers
asked Dec 14 '18 at 22:28
user398843user398843
675216
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$begingroup$
Theorem. Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0) = frac{1}{2pi i} int_C frac{f(z)dz}{z-z_0}.$$
The proof starts with letting $C_r$ denote a positively oriented circle $|z-z_0|=r$, where $r$ is small enough that $C_r$ is interior to $C$. Since the function $f(z)/(z-z_0)$ is analytic between and on the contours $C$ and $C_r$, we have $$ int_C frac{f(z)dz}{z-z_0} = int_{C_r} frac{f(z)dz}{z-z_0}.$$
Can we then parametrize the contour $C_r$ and calculate $int_{C_r} frac{f(z)dz}{z-z_0}$? How to proceed?
calculus complex-numbers
asked Dec 14 '18 at 22:28
user398843user398843
675216
$endgroup$
add a comment |
$begingroup$
Theorem. Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0) = frac{1}{2pi i} int_C frac{f(z)dz}{z-z_0}.$$
The proof starts with letting $C_r$ denote a positively oriented circle $|z-z_0|=r$, where $r$ is small enough that $C_r$ is interior to $C$. Since the function $f(z)/(z-z_0)$ is analytic between and on the contours $C$ and $C_r$, we have $$ int_C frac{f(z)dz}{z-z_0} = int_{C_r} frac{f(z)dz}{z-z_0}.$$
Can we then parametrize the contour $C_r$ and calculate $int_{C_r} frac{f(z)dz}{z-z_0}$? How to proceed?
calculus complex-numbers
asked Dec 14 '18 at 22:28
user398843user398843
675216
$endgroup$
Theorem. Let $f$ be analytic everywhere inside and on a simple closed contour $C$, taken in the positive sense. If $z_0$ is any point interior to $C$, then $$f(z_0) = frac{1}{2pi i} int_C frac{f(z)dz}{z-z_0}.$$
The proof starts with letting $C_r$ denote a positively oriented circle $|z-z_0|=r$, where $r$ is small enough that $C_r$ is interior to $C$. Since the function $f(z)/(z-z_0)$ is analytic between and on the contours $C$ and $C_r$, we have $$ int_C frac{f(z)dz}{z-z_0} = int_{C_r} frac{f(z)dz}{z-z_0}.$$
Can we then parametrize the contour $C_r$ and calculate $int_{C_r} frac{f(z)dz}{z-z_0}$? How to proceed?
calculus complex-numbers
calculus complex-numbers
asked Dec 14 '18 at 22:28
user398843user398843
675216
asked Dec 14 '18 at 22:28
user398843user398843
675216
edited Dec 14 '18 at 22:35
user398843
asked Dec 14 '18 at 22:28
user398843user398843
675216
asked Dec 14 '18 at 22:28
user398843user398843
675216
asked Dec 14 '18 at 22:28
user398843user398843
675216
675216
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2 Answers
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$begingroup$
One can proceed using mean value theorem and $rto 0$:
begin{align}
int_{C_r}frac{f(z)}{z-z_0},dz&=int_0^{2pi}frac{f(z_0+re^{itheta})}{re^{itheta}}re^{itheta}i,dtheta=iint_0^{2pi}f(z_0+re^{itheta}),dtheta=\
&=[text{Re},f(z_0+re^{ic_1})+itext{Im},f(z_0+re^{ic_2})]2pi ito f(z_0)2pi i.
end{align}
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
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$f(z)=sum a_n (z-z_0)^{n}$ gives $int_0^{2pi} f(z+re^{itheta}) dtheta=sum a_n int_0^{2pi}r^{n} e^{intheta} dtheta=2 pi a_0=2 pi f(z_0)$ (since all but the first term vanish). Now $int_{C_r} frac {f(z)} {z-z_0} dz =i int_0^{2pi} f(z+re^{itheta}) dtheta=2pi i f(z_0)$.
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
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2 Answers
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2 Answers
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active
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$begingroup$
One can proceed using mean value theorem and $rto 0$:
begin{align}
int_{C_r}frac{f(z)}{z-z_0},dz&=int_0^{2pi}frac{f(z_0+re^{itheta})}{re^{itheta}}re^{itheta}i,dtheta=iint_0^{2pi}f(z_0+re^{itheta}),dtheta=\
&=[text{Re},f(z_0+re^{ic_1})+itext{Im},f(z_0+re^{ic_2})]2pi ito f(z_0)2pi i.
end{align}
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
One can proceed using mean value theorem and $rto 0$:
begin{align}
int_{C_r}frac{f(z)}{z-z_0},dz&=int_0^{2pi}frac{f(z_0+re^{itheta})}{re^{itheta}}re^{itheta}i,dtheta=iint_0^{2pi}f(z_0+re^{itheta}),dtheta=\
&=[text{Re},f(z_0+re^{ic_1})+itext{Im},f(z_0+re^{ic_2})]2pi ito f(z_0)2pi i.
end{align}
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
One can proceed using mean value theorem and $rto 0$:
begin{align}
int_{C_r}frac{f(z)}{z-z_0},dz&=int_0^{2pi}frac{f(z_0+re^{itheta})}{re^{itheta}}re^{itheta}i,dtheta=iint_0^{2pi}f(z_0+re^{itheta}),dtheta=\
&=[text{Re},f(z_0+re^{ic_1})+itext{Im},f(z_0+re^{ic_2})]2pi ito f(z_0)2pi i.
end{align}
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
$endgroup$
One can proceed using mean value theorem and $rto 0$:
begin{align}
int_{C_r}frac{f(z)}{z-z_0},dz&=int_0^{2pi}frac{f(z_0+re^{itheta})}{re^{itheta}}re^{itheta}i,dtheta=iint_0^{2pi}f(z_0+re^{itheta}),dtheta=\
&=[text{Re},f(z_0+re^{ic_1})+itext{Im},f(z_0+re^{ic_2})]2pi ito f(z_0)2pi i.
end{align}
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
edited Dec 14 '18 at 23:33
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
answered Dec 14 '18 at 23:23
A.Γ.A.Γ.
22.8k32656
22.8k32656
add a comment |
add a comment |
$begingroup$
$f(z)=sum a_n (z-z_0)^{n}$ gives $int_0^{2pi} f(z+re^{itheta}) dtheta=sum a_n int_0^{2pi}r^{n} e^{intheta} dtheta=2 pi a_0=2 pi f(z_0)$ (since all but the first term vanish). Now $int_{C_r} frac {f(z)} {z-z_0} dz =i int_0^{2pi} f(z+re^{itheta}) dtheta=2pi i f(z_0)$.
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
add a comment |
$begingroup$
$f(z)=sum a_n (z-z_0)^{n}$ gives $int_0^{2pi} f(z+re^{itheta}) dtheta=sum a_n int_0^{2pi}r^{n} e^{intheta} dtheta=2 pi a_0=2 pi f(z_0)$ (since all but the first term vanish). Now $int_{C_r} frac {f(z)} {z-z_0} dz =i int_0^{2pi} f(z+re^{itheta}) dtheta=2pi i f(z_0)$.
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
add a comment |
$begingroup$
$f(z)=sum a_n (z-z_0)^{n}$ gives $int_0^{2pi} f(z+re^{itheta}) dtheta=sum a_n int_0^{2pi}r^{n} e^{intheta} dtheta=2 pi a_0=2 pi f(z_0)$ (since all but the first term vanish). Now $int_{C_r} frac {f(z)} {z-z_0} dz =i int_0^{2pi} f(z+re^{itheta}) dtheta=2pi i f(z_0)$.
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
$f(z)=sum a_n (z-z_0)^{n}$ gives $int_0^{2pi} f(z+re^{itheta}) dtheta=sum a_n int_0^{2pi}r^{n} e^{intheta} dtheta=2 pi a_0=2 pi f(z_0)$ (since all but the first term vanish). Now $int_{C_r} frac {f(z)} {z-z_0} dz =i int_0^{2pi} f(z+re^{itheta}) dtheta=2pi i f(z_0)$.
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Dec 14 '18 at 23:23
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
65.1k42766
add a comment |
add a comment |
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