Mixing
Solve $int_{|z| = 3} tan (pi z) dz$ using argument principle
2
$begingroup$
$int_{|z| = 3} tan (pi z) dz = int frac{d(cos pi z)}{ (-pi)cos pi z} dz = -2i(N-P)$
where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)
zeroes for $cos pi z$ = -0.5,-1.5,-2.5,0.5,1.5,2.5 $implies $ N = 6
No poles for $cos pi z$
value of given integral = $-2i (N-P) = -2i(6-0) = -12i$
Is this correct? pls correct me if i am doing wrong
complex-analysis trigonometry contour-integration
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
$endgroup$
add a comment |
2
$begingroup$
$int_{|z| = 3} tan (pi z) dz = int frac{d(cos pi z)}{ (-pi)cos pi z} dz = -2i(N-P)$
where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)
zeroes for $cos pi z$ = -0.5,-1.5,-2.5,0.5,1.5,2.5 $implies $ N = 6
No poles for $cos pi z$
value of given integral = $-2i (N-P) = -2i(6-0) = -12i$
Is this correct? pls correct me if i am doing wrong
complex-analysis trigonometry contour-integration
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
$endgroup$
$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15
add a comment |
2
2
2
$begingroup$
$int_{|z| = 3} tan (pi z) dz = int frac{d(cos pi z)}{ (-pi)cos pi z} dz = -2i(N-P)$
where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)
zeroes for $cos pi z$ = -0.5,-1.5,-2.5,0.5,1.5,2.5 $implies $ N = 6
No poles for $cos pi z$
value of given integral = $-2i (N-P) = -2i(6-0) = -12i$
Is this correct? pls correct me if i am doing wrong
complex-analysis trigonometry contour-integration
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
$endgroup$
$int_{|z| = 3} tan (pi z) dz = int frac{d(cos pi z)}{ (-pi)cos pi z} dz = -2i(N-P)$
where N = num of zeroes inside C:|z| = 3 and P is num of poles inside C (Is this correct or should we also consider on C???)
zeroes for $cos pi z$ = -0.5,-1.5,-2.5,0.5,1.5,2.5 $implies $ N = 6
No poles for $cos pi z$
value of given integral = $-2i (N-P) = -2i(6-0) = -12i$
Is this correct? pls correct me if i am doing wrong
complex-analysis trigonometry contour-integration
complex-analysis trigonometry contour-integration
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
edited Jan 18 at 11:23
José Carlos Santos
164k22131234
164k22131234
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
asked Jun 9 '18 at 18:56
MagnetoMagneto
881213
881213
$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15
add a comment |
$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15
$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $int_Cfrac{f'(z)}{f(z)},mathrm dz$ would make no sense.
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
|
show 1 more comment
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $int_Cfrac{f'(z)}{f(z)},mathrm dz$ would make no sense.
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
|
show 1 more comment
3
$begingroup$
It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $int_Cfrac{f'(z)}{f(z)},mathrm dz$ would make no sense.
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
|
show 1 more comment
3
3
3
$begingroup$
It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $int_Cfrac{f'(z)}{f(z)},mathrm dz$ would make no sense.
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
It is correct, yes. And the zeros and polse are inside $C$. Actually, if there were zeros or poles on $C$, the expressin $int_Cfrac{f'(z)}{f(z)},mathrm dz$ would make no sense.
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
answered Jun 9 '18 at 19:06
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
|
show 1 more comment
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
how can we conclude that it does not make sense? is this theorem itself is related to "inside C" or any logical way to conclude? pls elaborate.
$endgroup$
– Magneto
Jun 9 '18 at 19:10
3
3
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
@anirudhb The integrand is not a continuous complex-valued on $C$ if $f$ has a pole on $C$. The theorem is for inside $C$: are you asking for a proof?
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:14
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
$begingroup$
No, not proof. I am just thinking what happens to integral when we consider ON C. Will it breaks?
$endgroup$
– Magneto
Jun 9 '18 at 19:18
1
1
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
@anirudhb The integral will not make sense then because the function $frac{f'}f$ will not be defined in certain points of the path $C$.
$endgroup$
– José Carlos Santos
Jun 9 '18 at 19:20
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
$begingroup$
Oh, yes.... i got it. My brain got dumb... if f = 0 on C then 1/f goes to infinity. Thank you so much for patiently explaining me
$endgroup$
– Magneto
Jun 9 '18 at 19:22
|
show 1 more comment
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$begingroup$
The argument principle counts zeroes and poles with multiplicity. In this case, all zeroes and poles have multiplicity one (why?).
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:09
$begingroup$
cos pi z has no poles and only zeroes at +-2.5,+-1.5,+-0.5 --> zeroes of multiplicity 1
$endgroup$
– Magneto
Jun 9 '18 at 19:10
$begingroup$
$f:mathbb{C} to mathbb{C}$ by $f(z) = cos(pi z)^2$ has the same set of zeroes and poles, but the zeroes have different multiplicities. To argue that the zeroes are multiplicity 1, you have to check to see whether $frac{d}{dz} cos(pi z)$ vanishes.
$endgroup$
– Joshua Mundinger
Jun 9 '18 at 19:13
$begingroup$
here $f(z) = cos pi z$ and so it becomes given integral expression. for $cos pi z$, there are no poles only zeroes inside C. There is no $cos (pi z)^2$
$endgroup$
– Magneto
Jun 9 '18 at 19:15