Compute \[ \int_\gamma {z^2+1 \over z(z^2 + 4)}\dz \] for
- A circle of radius \(1\),
- A circle of radius \(2+\eps\),
Solution
For the smaller circle, use Cauchy’s formula \[ \int_{\abs{z} = 1} f(z) \dz &= \int_{\abs{z} = 1} { {z^2 + 1 \over z^2 + 4} \over z} \dz \\ &= 2\pi i \qty{z^2 + 1 \over z^2 + 4} \evalfrom_{z=0} \\ &= {i\pi \over 2} .\].
For the larger circle, break into principal parts. Write \[ {z^2 + 1 \over z(z^2+4)} = {a\over z} + {b\over z+2i} + {c\over z-2i} ,\] and use PFD/cover-up method/finding residues:
- \(a = \qty{z^2+1\over z^2+4}\evalfrom_{z=0} = {1\over 4}\)
- \(b= \qty{z^2 + 1 \over z(z-2i)}\evalfrom_{z=-2i} = {-4+1\over -2i(-4i)} = {3\over 8}\)
- \(c = \qty{z^2 + 1 \over z(z+2i)}\evalfrom_{z=2i} = {-4+1 \over 2i(4i)} = {3\over 8}\)
Thus \[ \int_\gamma f(z) \dz &= \int_\gamma \qty{ {1/4\over z} + {3/8\over z+2i} + {3/8 \over z-2i} } \dz \\ &= 2\pi i\qty{ {1\over 4} + {3\over 8} + {3\over 8}} .\]