Use Cauchy’s Integral formulas to evaluate the following integral along the
indicated closed contours.
∮
z
2+3z+2i
z
2+3z−4
dz;
C
(a) |z| = 2 (b) |z + 5| =
3
2
Cauchy’s integral formula:
"f(z_0)=\\frac{1}{2\\pi i}\\int_C \\frac{f(z)}{z-z_0}dz"
for any z0 inside C
we have:
"\\int_C \\frac{z^2+3z+2i}{z^2+3z-4}dz"
"z^2+3z-4=0"
"z=\\frac{-3\\pm \\sqrt{9+16}}{2}"
so,
"z_0=-4,1"
"\\frac{z^2+3z+2i}{z^2+3z-4}=\\frac{z^2+3z+2i}{(z+4)(z-1)}"
a)
for |z| = 2:
"z_0=1" is inside C
then:
"f(z)=\\frac{z^2+3z+2i}{z+4}"
"\\int_C \\frac{z^2+3z+2i}{z^2+3z-4}dz=2\\pi if(1)=2\\pi i(4+2i)\/5=\\pi(-4 +8i)\/5"
b)
for |z + 5| = 3/2:
"z_0=-4" is inside C
then:
"f(z)=\\frac{z^2+3z+2i}{z-1}"
"\\int_C \\frac{z^2+3z+2i}{z^2+3z-4}dz=2\\pi if(-4)=-2\\pi i(4+2i)\/5=\\pi(4 -8i)\/5"
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