Mixing
Can someone help me? I started but i have problem
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$int_Φ frac{e^{2z} } {z^4(z-2)}dz$ where Φ:[0,2Pi] and $Φ(t)=3e^{it} $ so we have $z=3e^{it}$ and $dz=3ie^{it}dt$ I want to calculate $ int_0^{2Pi} frac{e^{6e^{it}}} {e^{4}(3e^{it}-2)} 3ie^{it}dt = 3i int_0^{2Pi} frac{e^{7e^{it}}} {e^{4}(3e^{it}-2)} dt=3i int_0^{2Pi} frac{e^{7e^{it}}} {3e^{4it}-2e^{4}} dt=
3ie^{-4} int_0^{2Pi} frac{e^{7e^{it}}e^{-it}} {3-2e^{-it}} dt$ and now i take $e^{-it} =w$ and $-ie^{-it} dt=dw$ so we have $frac{-3i} {ie^4}int_0^{2Pi} frac{e^{7w^{-1}}dw} {3-2w} dw$,and I dont know what should I do now
complex-analysis contour-integration
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
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add a comment |
$begingroup$
$int_Φ frac{e^{2z} } {z^4(z-2)}dz$ where Φ:[0,2Pi] and $Φ(t)=3e^{it} $ so we have $z=3e^{it}$ and $dz=3ie^{it}dt$ I want to calculate $ int_0^{2Pi} frac{e^{6e^{it}}} {e^{4}(3e^{it}-2)} 3ie^{it}dt = 3i int_0^{2Pi} frac{e^{7e^{it}}} {e^{4}(3e^{it}-2)} dt=3i int_0^{2Pi} frac{e^{7e^{it}}} {3e^{4it}-2e^{4}} dt=
3ie^{-4} int_0^{2Pi} frac{e^{7e^{it}}e^{-it}} {3-2e^{-it}} dt$ and now i take $e^{-it} =w$ and $-ie^{-it} dt=dw$ so we have $frac{-3i} {ie^4}int_0^{2Pi} frac{e^{7w^{-1}}dw} {3-2w} dw$,and I dont know what should I do now
complex-analysis contour-integration
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
$endgroup$
add a comment |
$begingroup$
$int_Φ frac{e^{2z} } {z^4(z-2)}dz$ where Φ:[0,2Pi] and $Φ(t)=3e^{it} $ so we have $z=3e^{it}$ and $dz=3ie^{it}dt$ I want to calculate $ int_0^{2Pi} frac{e^{6e^{it}}} {e^{4}(3e^{it}-2)} 3ie^{it}dt = 3i int_0^{2Pi} frac{e^{7e^{it}}} {e^{4}(3e^{it}-2)} dt=3i int_0^{2Pi} frac{e^{7e^{it}}} {3e^{4it}-2e^{4}} dt=
3ie^{-4} int_0^{2Pi} frac{e^{7e^{it}}e^{-it}} {3-2e^{-it}} dt$ and now i take $e^{-it} =w$ and $-ie^{-it} dt=dw$ so we have $frac{-3i} {ie^4}int_0^{2Pi} frac{e^{7w^{-1}}dw} {3-2w} dw$,and I dont know what should I do now
complex-analysis contour-integration
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
$endgroup$
$int_Φ frac{e^{2z} } {z^4(z-2)}dz$ where Φ:[0,2Pi] and $Φ(t)=3e^{it} $ so we have $z=3e^{it}$ and $dz=3ie^{it}dt$ I want to calculate $ int_0^{2Pi} frac{e^{6e^{it}}} {e^{4}(3e^{it}-2)} 3ie^{it}dt = 3i int_0^{2Pi} frac{e^{7e^{it}}} {e^{4}(3e^{it}-2)} dt=3i int_0^{2Pi} frac{e^{7e^{it}}} {3e^{4it}-2e^{4}} dt=
3ie^{-4} int_0^{2Pi} frac{e^{7e^{it}}e^{-it}} {3-2e^{-it}} dt$ and now i take $e^{-it} =w$ and $-ie^{-it} dt=dw$ so we have $frac{-3i} {ie^4}int_0^{2Pi} frac{e^{7w^{-1}}dw} {3-2w} dw$,and I dont know what should I do now
complex-analysis contour-integration
complex-analysis contour-integration
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
edited Jan 19 at 15:49
José Carlos Santos
164k22132235
164k22132235
asked Jan 19 at 15:28
KamiloKamilo
43
asked Jan 19 at 15:28
KamiloKamilo
43
asked Jan 19 at 15:28
KamiloKamilo
43
43
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
You can use the fact that$$frac{e^{2z}}{z^4(z-2)}=frac{e^{2z}}{16(z-2)}-frac{e^{2z}}{2z^4}-frac{e^{2z}}{4z^3}-frac{e^{2z}}{8z^2}-frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
add a comment |
$begingroup$
Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
$begingroup$
You can use the fact that$$frac{e^{2z}}{z^4(z-2)}=frac{e^{2z}}{16(z-2)}-frac{e^{2z}}{2z^4}-frac{e^{2z}}{4z^3}-frac{e^{2z}}{8z^2}-frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
add a comment |
$begingroup$
You can use the fact that$$frac{e^{2z}}{z^4(z-2)}=frac{e^{2z}}{16(z-2)}-frac{e^{2z}}{2z^4}-frac{e^{2z}}{4z^3}-frac{e^{2z}}{8z^2}-frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
add a comment |
$begingroup$
You can use the fact that$$frac{e^{2z}}{z^4(z-2)}=frac{e^{2z}}{16(z-2)}-frac{e^{2z}}{2z^4}-frac{e^{2z}}{4z^3}-frac{e^{2z}}{8z^2}-frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
You can use the fact that$$frac{e^{2z}}{z^4(z-2)}=frac{e^{2z}}{16(z-2)}-frac{e^{2z}}{2z^4}-frac{e^{2z}}{4z^3}-frac{e^{2z}}{8z^2}-frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 15:36
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
add a comment |
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
Thanks! Now is easy :) but how you created this formula? Is this any rule?
$endgroup$
– Kamilo
Jan 19 at 17:15
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
I applied partial fraction decomposition to $frac1{z^4(z-2)}$.
$endgroup$
– José Carlos Santos
Jan 19 at 17:18
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
$begingroup$
Thanks! This was very useful! 😁
$endgroup$
– Kamilo
Jan 19 at 17:48
add a comment |
$begingroup$
Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
add a comment |
$begingroup$
Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
add a comment |
$begingroup$
Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
$endgroup$
Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
edited Jan 19 at 15:31
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
answered Jan 19 at 15:30
Tsemo AristideTsemo Aristide
58.9k11445
58.9k11445
add a comment |
add a comment |
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