Evaluate $frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$Compute $int_{gamma}...
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Evaluate $frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$
Compute $int_{gamma} frac{z}{z^3-1} dz$ where $gamma$ is circle centered at origin of radius 2Find $int_Gammafrac{3z-2}{z^2-z}dz$, where $Gamma$ encloses point $0$ and $1$.If three complex numbers $z_k$ have modulus $1$, then $|z_1+z_2+z_3| = |frac{1}{z_1}+frac{1}{z_2}+frac{1}{z_3}|$Evaluate $int_{|C|=2} frac{dz}{z^2 + 2z + 2}$ using Cauchy-GoursatEvaluate $ int_{gamma} (z + frac {1}{z - 1}) , dz $ where $ gamma $ is the perimeter of the parallelogram with vertices $ i, -i, 2 + i, 2 - i $Cauchy Integral Formula and Fundamental Theorem of AlgebraShow that $int_Gamma frac{f'(z)}{z-z_0}dz=int_Gammafrac{f(z)}{(z-z_0)^2}dz$Assuming that $g(z)=frac{f(z)}{z-z_0}$ is continuous at $z_0$, prove that $int_gamma frac{f(z)}{z-z_0}dz=0$.a function that is analytic which satisfies a Lipschitz conditionUsing Cauchy theorem to compute $int_{gamma} zoverline{z}$
$begingroup$
In Marsden's Complex Analysis, section 2.4, the main theorem is Cauchy's integral formula (C.I.F) and there appears this problem:
Let $f$ be analytic inside and on $gamma: |z-z_0|=R$. Prove that
$$frac{f(z_1)-f(z_2)}{z_1-z_2}-f'(z_0)=frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$$
for $z_1, z_2$ inside $gamma$
My attempt:
That $$-f'(z_0) = - frac{1}{2 pi i} int_gamma frac{f(z)}{(z-z_0)^2}$$
follows from Cauchy's integral formula.
In order to evaluate $$frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}$$
we can draw two "very" small circles, parametrized as curves oriented negatively, around $z_1$ and $z_2$ respectively. I called them $- alpha$ and $- beta$. It can be shown that
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -frac{1}{2pi i}int_alphafrac{f(z)}{(z-z_1)(z-z_2)} - frac{1}{2pi i}int_betafrac{f(z)}{(z-z_1)(z-z_2)} $$
(to see this, according to figure draw a line connecting $z_1$ and $z_2$, dividing the greatest circle $gamma$ in two regions, A and B, where $frac{f(z)}{(z-z_1)(z-z_2)}$ is analytic and hence the integral es $0$ trough the frontier in each of these regions (the curves with arrows blue and red in the figure) and hence the integral is $0$ in the sum of the two curves. And after canceling the segments of line (because in the red curve are traveled in opposite direction with respect of the blue curve), we get our claim (the small remaining circles around $z_1$ and $z_2$ are $alpha$ and $beta$))
But $frac{f(z)}{(z-z_2)}$ is analytic inside $alpha$ and $frac{f(z)}{(z-z_1)}$ is analytic inside $beta$. So Cauchy's formula gives
$$frac{1}{2 pi i}int_alpha frac{f(z)}{(z-z_2)(z-z_1)} = frac{f(z_1)}{(z_1-z_2)} = -frac{f(z_1)}{(z_2-z_1)}$$
and
$$frac{1}{2 pi i}int_beta frac{f(z)}{(z-z_1)(z-z_2)} = frac{f(z_2)}{(z_2-z_1)}= frac{f(z_2)}{(z_2-z_1)} $$
hence $$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -left(frac{f(z_2)-f(z_1)}{(z_2-z_1)} right) = frac{f(z_1)-f(z_2)}{(z_2-z_1)}$$
So I get the desired result, except that $f(z_1)$ is interchanged with $f(z_2)$. Where do I committed a mistake?
proof-verification complex-integration cauchy-integral-formula winding-number
asked Mar 20 at 0:47
seferpdseferpd
85
$endgroup$
add a comment |
$begingroup$
In Marsden's Complex Analysis, section 2.4, the main theorem is Cauchy's integral formula (C.I.F) and there appears this problem:
Let $f$ be analytic inside and on $gamma: |z-z_0|=R$. Prove that
$$frac{f(z_1)-f(z_2)}{z_1-z_2}-f'(z_0)=frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$$
for $z_1, z_2$ inside $gamma$
My attempt:
That $$-f'(z_0) = - frac{1}{2 pi i} int_gamma frac{f(z)}{(z-z_0)^2}$$
follows from Cauchy's integral formula.
In order to evaluate $$frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}$$
we can draw two "very" small circles, parametrized as curves oriented negatively, around $z_1$ and $z_2$ respectively. I called them $- alpha$ and $- beta$. It can be shown that
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -frac{1}{2pi i}int_alphafrac{f(z)}{(z-z_1)(z-z_2)} - frac{1}{2pi i}int_betafrac{f(z)}{(z-z_1)(z-z_2)} $$
(to see this, according to figure draw a line connecting $z_1$ and $z_2$, dividing the greatest circle $gamma$ in two regions, A and B, where $frac{f(z)}{(z-z_1)(z-z_2)}$ is analytic and hence the integral es $0$ trough the frontier in each of these regions (the curves with arrows blue and red in the figure) and hence the integral is $0$ in the sum of the two curves. And after canceling the segments of line (because in the red curve are traveled in opposite direction with respect of the blue curve), we get our claim (the small remaining circles around $z_1$ and $z_2$ are $alpha$ and $beta$))
But $frac{f(z)}{(z-z_2)}$ is analytic inside $alpha$ and $frac{f(z)}{(z-z_1)}$ is analytic inside $beta$. So Cauchy's formula gives
$$frac{1}{2 pi i}int_alpha frac{f(z)}{(z-z_2)(z-z_1)} = frac{f(z_1)}{(z_1-z_2)} = -frac{f(z_1)}{(z_2-z_1)}$$
and
$$frac{1}{2 pi i}int_beta frac{f(z)}{(z-z_1)(z-z_2)} = frac{f(z_2)}{(z_2-z_1)}= frac{f(z_2)}{(z_2-z_1)} $$
hence $$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -left(frac{f(z_2)-f(z_1)}{(z_2-z_1)} right) = frac{f(z_1)-f(z_2)}{(z_2-z_1)}$$
So I get the desired result, except that $f(z_1)$ is interchanged with $f(z_2)$. Where do I committed a mistake?
proof-verification complex-integration cauchy-integral-formula winding-number
asked Mar 20 at 0:47
seferpdseferpd
85
$endgroup$
add a comment |
$begingroup$
In Marsden's Complex Analysis, section 2.4, the main theorem is Cauchy's integral formula (C.I.F) and there appears this problem:
Let $f$ be analytic inside and on $gamma: |z-z_0|=R$. Prove that
$$frac{f(z_1)-f(z_2)}{z_1-z_2}-f'(z_0)=frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$$
for $z_1, z_2$ inside $gamma$
My attempt:
That $$-f'(z_0) = - frac{1}{2 pi i} int_gamma frac{f(z)}{(z-z_0)^2}$$
follows from Cauchy's integral formula.
In order to evaluate $$frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}$$
we can draw two "very" small circles, parametrized as curves oriented negatively, around $z_1$ and $z_2$ respectively. I called them $- alpha$ and $- beta$. It can be shown that
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -frac{1}{2pi i}int_alphafrac{f(z)}{(z-z_1)(z-z_2)} - frac{1}{2pi i}int_betafrac{f(z)}{(z-z_1)(z-z_2)} $$
(to see this, according to figure draw a line connecting $z_1$ and $z_2$, dividing the greatest circle $gamma$ in two regions, A and B, where $frac{f(z)}{(z-z_1)(z-z_2)}$ is analytic and hence the integral es $0$ trough the frontier in each of these regions (the curves with arrows blue and red in the figure) and hence the integral is $0$ in the sum of the two curves. And after canceling the segments of line (because in the red curve are traveled in opposite direction with respect of the blue curve), we get our claim (the small remaining circles around $z_1$ and $z_2$ are $alpha$ and $beta$))
But $frac{f(z)}{(z-z_2)}$ is analytic inside $alpha$ and $frac{f(z)}{(z-z_1)}$ is analytic inside $beta$. So Cauchy's formula gives
$$frac{1}{2 pi i}int_alpha frac{f(z)}{(z-z_2)(z-z_1)} = frac{f(z_1)}{(z_1-z_2)} = -frac{f(z_1)}{(z_2-z_1)}$$
and
$$frac{1}{2 pi i}int_beta frac{f(z)}{(z-z_1)(z-z_2)} = frac{f(z_2)}{(z_2-z_1)}= frac{f(z_2)}{(z_2-z_1)} $$
hence $$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -left(frac{f(z_2)-f(z_1)}{(z_2-z_1)} right) = frac{f(z_1)-f(z_2)}{(z_2-z_1)}$$
So I get the desired result, except that $f(z_1)$ is interchanged with $f(z_2)$. Where do I committed a mistake?
proof-verification complex-integration cauchy-integral-formula winding-number
asked Mar 20 at 0:47
seferpdseferpd
85
$endgroup$
In Marsden's Complex Analysis, section 2.4, the main theorem is Cauchy's integral formula (C.I.F) and there appears this problem:
Let $f$ be analytic inside and on $gamma: |z-z_0|=R$. Prove that
$$frac{f(z_1)-f(z_2)}{z_1-z_2}-f'(z_0)=frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}-frac{f(z)}{(z-z_0)^2}$$
for $z_1, z_2$ inside $gamma$
My attempt:
That $$-f'(z_0) = - frac{1}{2 pi i} int_gamma frac{f(z)}{(z-z_0)^2}$$
follows from Cauchy's integral formula.
In order to evaluate $$frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}$$
we can draw two "very" small circles, parametrized as curves oriented negatively, around $z_1$ and $z_2$ respectively. I called them $- alpha$ and $- beta$. It can be shown that
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -frac{1}{2pi i}int_alphafrac{f(z)}{(z-z_1)(z-z_2)} - frac{1}{2pi i}int_betafrac{f(z)}{(z-z_1)(z-z_2)} $$
(to see this, according to figure draw a line connecting $z_1$ and $z_2$, dividing the greatest circle $gamma$ in two regions, A and B, where $frac{f(z)}{(z-z_1)(z-z_2)}$ is analytic and hence the integral es $0$ trough the frontier in each of these regions (the curves with arrows blue and red in the figure) and hence the integral is $0$ in the sum of the two curves. And after canceling the segments of line (because in the red curve are traveled in opposite direction with respect of the blue curve), we get our claim (the small remaining circles around $z_1$ and $z_2$ are $alpha$ and $beta$))
But $frac{f(z)}{(z-z_2)}$ is analytic inside $alpha$ and $frac{f(z)}{(z-z_1)}$ is analytic inside $beta$. So Cauchy's formula gives
$$frac{1}{2 pi i}int_alpha frac{f(z)}{(z-z_2)(z-z_1)} = frac{f(z_1)}{(z_1-z_2)} = -frac{f(z_1)}{(z_2-z_1)}$$
and
$$frac{1}{2 pi i}int_beta frac{f(z)}{(z-z_1)(z-z_2)} = frac{f(z_2)}{(z_2-z_1)}= frac{f(z_2)}{(z_2-z_1)} $$
hence $$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)} = -left(frac{f(z_2)-f(z_1)}{(z_2-z_1)} right) = frac{f(z_1)-f(z_2)}{(z_2-z_1)}$$
So I get the desired result, except that $f(z_1)$ is interchanged with $f(z_2)$. Where do I committed a mistake?
proof-verification complex-integration cauchy-integral-formula winding-number
proof-verification complex-integration cauchy-integral-formula winding-number
asked Mar 20 at 0:47
seferpdseferpd
85
asked Mar 20 at 0:47
seferpdseferpd
85
edited Mar 20 at 0:57
seferpd
asked Mar 20 at 0:47
seferpdseferpd
85
asked Mar 20 at 0:47
seferpdseferpd
85
asked Mar 20 at 0:47
seferpdseferpd
85
85
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint: using
$$ frac{1}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}left(frac{1}{z-z_1}-frac{1}{z-z_2}right) $$
and Cauchy's Integral Formula, then one has
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}frac{1}{2pi i}int_gammaleft(frac{f(z)}{z-z_1}-frac{f(z)}{z-z_2}right)=frac{1}{z_1-z_2}(f(z_1)-f(z_2)). $$
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: using
$$ frac{1}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}left(frac{1}{z-z_1}-frac{1}{z-z_2}right) $$
and Cauchy's Integral Formula, then one has
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}frac{1}{2pi i}int_gammaleft(frac{f(z)}{z-z_1}-frac{f(z)}{z-z_2}right)=frac{1}{z_1-z_2}(f(z_1)-f(z_2)). $$
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
$begingroup$
Hint: using
$$ frac{1}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}left(frac{1}{z-z_1}-frac{1}{z-z_2}right) $$
and Cauchy's Integral Formula, then one has
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}frac{1}{2pi i}int_gammaleft(frac{f(z)}{z-z_1}-frac{f(z)}{z-z_2}right)=frac{1}{z_1-z_2}(f(z_1)-f(z_2)). $$
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
$endgroup$
add a comment |
$begingroup$
Hint: using
$$ frac{1}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}left(frac{1}{z-z_1}-frac{1}{z-z_2}right) $$
and Cauchy's Integral Formula, then one has
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}frac{1}{2pi i}int_gammaleft(frac{f(z)}{z-z_1}-frac{f(z)}{z-z_2}right)=frac{1}{z_1-z_2}(f(z_1)-f(z_2)). $$
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
$endgroup$
Hint: using
$$ frac{1}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}left(frac{1}{z-z_1}-frac{1}{z-z_2}right) $$
and Cauchy's Integral Formula, then one has
$$ frac{1}{2pi i}int_gammafrac{f(z)}{(z-z_1)(z-z_2)}=frac{1}{z_1-z_2}frac{1}{2pi i}int_gammaleft(frac{f(z)}{z-z_1}-frac{f(z)}{z-z_2}right)=frac{1}{z_1-z_2}(f(z_1)-f(z_2)). $$
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
edited Mar 20 at 1:06
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
answered Mar 20 at 1:00
xpaulxpaul
23.4k24655
23.4k24655
add a comment |
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