Calculate the contour integral for e^z/(z^2-2z-3) z=r for r=2 and r=4
I found the residue for Z1=-1 and Z3=3, they are REZ(f1,z1)= -1/(4e) and REZ(f2,z2)=(e^3)/4
Now i gotta calculate the integral, but how do i do that? I got a formula :
Integral(f(z)dz)=2*i*pi*[(REZ(f1,z1)+R… but that only applies if z1, z2 are on the r interval, what does that mean? :/ are z1,z2 outside r interval, is only one outside, are they both in? lol, please help me :D
I found the residue for Z1=-1 and Z3=3, they are REZ(f1,z1)= -1/(4e) and REZ(f2,z2)=(e^3)/4
Now i gotta calculate the integral, but how do i do that? I got a formula :
Integral(f(z)dz)=2*i*pi*[(REZ(f1,z1)+R… but that only applies if z1, z2 are on the r interval, what does that mean? :/ are z1,z2 outside r interval, is only one outside, are they both in? lol, please help me :D
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If I read the question correct then shouldn't the contour be the circle |z| = r for r=2 and r=4 (since z=r is just a point in the complex plane not a contour)? These are two separate contour integrals (i.e. one integral anticlockwise along the circle |z|=2 and another integral anticlockwise on a circle |z|=4 in the complex plane). If I take this as your meaning then calculate the value of your residues.
The first residue is for the singularity at z=-1 on the real line in the complex plane (i.e. coefficient of 1/(z+1) at the point z=z1=-1).
REZ(f1,z1) = -1/4e = -0.092 (1)
The second residue is for the singularity at z=3 on the real line in the complex plane (i.e. in this case the coefficient of 1/(z-3) at the point z=z2=3)
REZ(f2,z2) = (e^3)/4 = 5.02 (2)
z1=-1 lies inside both of the circles |z|=2 and |z|=4 but z2=3 lies only inside the circle |z|=4. So for the first contour |z|=2, when using your formula, include only the residue inside that circle (1) but for the second contour include both residues ((1) and (2)). If on the other hand I have misunderstood the question and you want to find the integral about both circles (i.e. anticlockwise about |z|=4 and clockwise about |z|=2) then only include the residue (2) since (1) is outside the region between the two circles.
In summary your answers should be:
Integral over contour anticlockwise around circle |z|=2:
Integral(f(z)dz) = 2ipi*REZ(f1,z1) = -0.58i (3)
Integral over contour anticlockwise around circle |z|=4:
Integral(f(z)dz) = 2ipi*[REZ(f1,z1)+REZ(f2,z2)] = -0.58i + 31.54i = 30.96i (4)
Integral over contour anticlockwise around circle |z|=4 and clockwise around circle |z|=2:
Integral(f(z)dz) = 2ipi*REZ(f2,z2) = 31.54i (5)
Note, when taking the last integral the path is actually two concentric circles joined by two adjacent and infintesimally close radial paths (one from circle |z|=2 to |z|=4 and the other in the opposite direction). These two paths are considered so close as to cancel one another.
Hope this has shed some light on a tricky subject!
The first residue is for the singularity at z=-1 on the real line in the complex plane (i.e. coefficient of 1/(z+1) at the point z=z1=-1).
REZ(f1,z1) = -1/4e = -0.092 (1)
The second residue is for the singularity at z=3 on the real line in the complex plane (i.e. in this case the coefficient of 1/(z-3) at the point z=z2=3)
REZ(f2,z2) = (e^3)/4 = 5.02 (2)
z1=-1 lies inside both of the circles |z|=2 and |z|=4 but z2=3 lies only inside the circle |z|=4. So for the first contour |z|=2, when using your formula, include only the residue inside that circle (1) but for the second contour include both residues ((1) and (2)). If on the other hand I have misunderstood the question and you want to find the integral about both circles (i.e. anticlockwise about |z|=4 and clockwise about |z|=2) then only include the residue (2) since (1) is outside the region between the two circles.
In summary your answers should be:
Integral over contour anticlockwise around circle |z|=2:
Integral(f(z)dz) = 2ipi*REZ(f1,z1) = -0.58i (3)
Integral over contour anticlockwise around circle |z|=4:
Integral(f(z)dz) = 2ipi*[REZ(f1,z1)+REZ(f2,z2)] = -0.58i + 31.54i = 30.96i (4)
Integral over contour anticlockwise around circle |z|=4 and clockwise around circle |z|=2:
Integral(f(z)dz) = 2ipi*REZ(f2,z2) = 31.54i (5)
Note, when taking the last integral the path is actually two concentric circles joined by two adjacent and infintesimally close radial paths (one from circle |z|=2 to |z|=4 and the other in the opposite direction). These two paths are considered so close as to cancel one another.
Hope this has shed some light on a tricky subject!