How to use Cauchy's integral formula with more than one pole?
$begingroup$
$$int_{gamma} frac{z^2}{z(z-2)}, quad gamma(theta) = 3e^{itheta}, 0 leq theta leq 2pi$$
Cauchy's integral formula is given by:
$$intlimits_{gamma} frac{f(z)}{(z-a)^{n+1}} = frac{2pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $f(z) = z^2$. But it doesn't seem like I can get my integral into a form like $(z - a)^n$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $gamma(theta)$ was $e^{itheta}$, then I could choose my holomorphic function to be $frac{z^2}{z-2}$?
integration complex-analysis contour-integration complex-integration
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
$endgroup$
add a comment |
$begingroup$
$$int_{gamma} frac{z^2}{z(z-2)}, quad gamma(theta) = 3e^{itheta}, 0 leq theta leq 2pi$$
Cauchy's integral formula is given by:
$$intlimits_{gamma} frac{f(z)}{(z-a)^{n+1}} = frac{2pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $f(z) = z^2$. But it doesn't seem like I can get my integral into a form like $(z - a)^n$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $gamma(theta)$ was $e^{itheta}$, then I could choose my holomorphic function to be $frac{z^2}{z-2}$?
integration complex-analysis contour-integration complex-integration
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
$endgroup$
$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
1
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
1
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40
add a comment |
$begingroup$
$$int_{gamma} frac{z^2}{z(z-2)}, quad gamma(theta) = 3e^{itheta}, 0 leq theta leq 2pi$$
Cauchy's integral formula is given by:
$$intlimits_{gamma} frac{f(z)}{(z-a)^{n+1}} = frac{2pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $f(z) = z^2$. But it doesn't seem like I can get my integral into a form like $(z - a)^n$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $gamma(theta)$ was $e^{itheta}$, then I could choose my holomorphic function to be $frac{z^2}{z-2}$?
integration complex-analysis contour-integration complex-integration
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
$endgroup$
$$int_{gamma} frac{z^2}{z(z-2)}, quad gamma(theta) = 3e^{itheta}, 0 leq theta leq 2pi$$
Cauchy's integral formula is given by:
$$intlimits_{gamma} frac{f(z)}{(z-a)^{n+1}} = frac{2pi i}{n!} f^{(n)}(a)$$
And I can choose my holomorphic $f(z) = z^2$. But it doesn't seem like I can get my integral into a form like $(z - a)^n$ in the denominator. Am I missing some algebraic trick to do this?
Also, if $gamma(theta)$ was $e^{itheta}$, then I could choose my holomorphic function to be $frac{z^2}{z-2}$?
integration complex-analysis contour-integration complex-integration
integration complex-analysis contour-integration complex-integration
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
edited Dec 30 '18 at 13:51
Lorenzo B.
1,8902520
1,8902520
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
asked Apr 17 '15 at 12:35
mr eyeglassesmr eyeglasses
2,48332248
2,48332248
$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
1
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
1
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40
add a comment |
$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
1
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
1
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40
$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
1
1
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
1
1
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40
add a comment |
1 Answer
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$begingroup$
We have
$$dfrac{z^2}{z(z-2)} = dfrac{z}{z-2} = 1 + dfrac2{z-2}$$
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We have
$$dfrac{z^2}{z(z-2)} = dfrac{z}{z-2} = 1 + dfrac2{z-2}$$
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
$endgroup$
add a comment |
$begingroup$
We have
$$dfrac{z^2}{z(z-2)} = dfrac{z}{z-2} = 1 + dfrac2{z-2}$$
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
$endgroup$
add a comment |
$begingroup$
We have
$$dfrac{z^2}{z(z-2)} = dfrac{z}{z-2} = 1 + dfrac2{z-2}$$
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
$endgroup$
We have
$$dfrac{z^2}{z(z-2)} = dfrac{z}{z-2} = 1 + dfrac2{z-2}$$
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
answered Apr 17 '15 at 12:42
LegLeg
18.7k11748
18.7k11748
add a comment |
add a comment |
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$begingroup$
en.m.wikipedia.org/wiki/Residue_theorem
$endgroup$
– Ant
Apr 17 '15 at 12:38
1
$begingroup$
Isn't $frac{z^2}{z(z-2)}=frac{z}{z-2}$? (with $z=0$ being a removal singularity)
$endgroup$
– kennytm
Apr 17 '15 at 12:39
1
$begingroup$
$frac{z^2}{zleft(z-2right)}=frac{z}{z-2}$ so that you can take $fleft(zright)=z$.
$endgroup$
– Nicolas
Apr 17 '15 at 12:40
$begingroup$
Oh, I didn't know we're allowed to cancel out fractions inside an integral. Thanks
$endgroup$
– mr eyeglasses
Apr 17 '15 at 12:40