Find a circle in which the integral sea different from zero
My function is $z^{2} hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
begin{equation}
int z^{2} hat z^{3} dx = int frac{64}{z} dx
end{equation}
This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
but I do not know if this is valid
integration complex-analysis
asked Nov 25 at 7:23
Jazmín Jones
519
add a comment |
My function is $z^{2} hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
begin{equation}
int z^{2} hat z^{3} dx = int frac{64}{z} dx
end{equation}
This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
but I do not know if this is valid
integration complex-analysis
asked Nov 25 at 7:23
Jazmín Jones
519
add a comment |
My function is $z^{2} hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
begin{equation}
int z^{2} hat z^{3} dx = int frac{64}{z} dx
end{equation}
This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
but I do not know if this is valid
integration complex-analysis
asked Nov 25 at 7:23
Jazmín Jones
519
My function is $z^{2} hat z^{3}$
I don't know if this can be:
Taking the radio circle 2
begin{equation}
int z^{2} hat z^{3} dx = int frac{64}{z} dx
end{equation}
This because $z hat z= |z|^2 = 4$ but this happens twice, so $z hat z z hat z= |z|^2 = 16$ y then there is a $hat z$ without $z$ so $z hat z = 4 rightarrow hat z = frac{4}{z}$
but I do not know if this is valid
integration complex-analysis
integration complex-analysis
asked Nov 25 at 7:23
Jazmín Jones
519
asked Nov 25 at 7:23
Jazmín Jones
519
asked Nov 25 at 7:23
Jazmín Jones
519
asked Nov 25 at 7:23
Jazmín Jones
519
asked Nov 25 at 7:23
Jazmín Jones
519
519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
$$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$
answered Nov 25 at 7:38
Guacho Perez
3,79711131
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
add a comment |
In your case the integrand is
$$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral
$$left.32iright|_0^{2pi}.$$
answered Nov 25 at 8:25
Yves Daoust
124k671220
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
$$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$
answered Nov 25 at 7:38
Guacho Perez
3,79711131
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
add a comment |
First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
$$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$
answered Nov 25 at 7:38
Guacho Perez
3,79711131
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
add a comment |
First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
$$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$
answered Nov 25 at 7:38
Guacho Perez
3,79711131
First, $z^2bar z^3=(zbar z)^2bar z=|z|^4bar z$ so using the unit circle $gamma=e^{it}$ gives
$$ int_{gamma}|z|^4bar z dz=int_0^{2pi} e^{-it}ie^{it}dt=2pi i$$
answered Nov 25 at 7:38
Guacho Perez
3,79711131
answered Nov 25 at 7:38
Guacho Perez
3,79711131
answered Nov 25 at 7:38
Guacho Perez
3,79711131
answered Nov 25 at 7:38
Guacho Perez
3,79711131
3,79711131
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
add a comment |
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
2
2
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
I guess @JazminJones mentioned "radius of circle being 2". So if thats the case $gamma=2e^{it}$ and the integral evaluates to $32pi i$.
– Yadati Kiran
Nov 25 at 7:47
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
$32cdot2pi i$ not $32pi i$.
– Yadati Kiran
Nov 25 at 8:37
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
can you explain the equality in more detail please
– Jazmín Jones
Nov 25 at 17:09
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
@JazminJones which one?
– Guacho Perez
Nov 25 at 18:06
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
Forget it thanks haha
– Jazmín Jones
Nov 25 at 18:14
add a comment |
In your case the integrand is
$$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral
$$left.32iright|_0^{2pi}.$$
answered Nov 25 at 8:25
Yves Daoust
124k671220
add a comment |
In your case the integrand is
$$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral
$$left.32iright|_0^{2pi}.$$
answered Nov 25 at 8:25
Yves Daoust
124k671220
add a comment |
In your case the integrand is
$$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral
$$left.32iright|_0^{2pi}.$$
answered Nov 25 at 8:25
Yves Daoust
124k671220
In your case the integrand is
$$(rho e^{itheta})^2(rho e^{-itheta})^3de^{itheta}=2^5e^{-itheta}de^{itheta}$$ giving the integral
$$left.32iright|_0^{2pi}.$$
answered Nov 25 at 8:25
Yves Daoust
124k671220
answered Nov 25 at 8:25
Yves Daoust
124k671220
answered Nov 25 at 8:25
Yves Daoust
124k671220
answered Nov 25 at 8:25
Yves Daoust
124k671220
124k671220
add a comment |
add a comment |
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