Evaluate the following integrals using Cauchy’s integral formula:
\[ \int_{S^1} {\cos(z) \over z} \dz&\\ \int_{S^1} {\sin(z) \over z}\dz&\\ \int_{\abs z = 2} {z^2\over z-1} \dz &\\ \int_{S^1} {e^z \over z^2} \dz &\\ \int_{\abs z = 2} {z^2 - 1 \over z^2 + 1}\dz &\\ \int_{\abs z = 2} {1\over z^2 + z + 1}\dz & .\]
Solution
\[ \int_{S^1} {\cos(z) \over z} \dz &= 2\pi i \cos(0) \\ &= 2\pi i \\ \\ \int_{S^1} {\sin(z) \over z}\dz &= 2\pi i \sin(0) \\ &= 0 \\ \\ \int_{\abs z = 2} {z^2\over z-1} \dz &= 2\pi i z^2\evalfrom_{z=1} \\ &= 2\pi i \\ \\ \int_{S^1} {e^z \over z^2} \dz &= 2\pi i \dd{}{z} e^z\evalfrom_{z=0} \\ &= 2\pi i \\ \\ \int_{\abs z = 2} {z^2 - 1 \over z^2 + 1}\dz &= \int_{\DD + i} {(z^2 -1)/(z+i) \over z-i}\dz + \int_{\DD - i} {(z^2-1)/(z-i) \over z+i} \dz \\ &= 2\pi i \qty{z^2-1\over z+i}\evalfrom_{z=i} + 2\pi i \qty{z^2-1 \over z-i}\evalfrom_{z=-i} \\ &= 2\pi i \qty{-2\over 2i} + 2\pi i \qty{-2\over -2i} \\ &= 0 \\ \\ \int_{\abs z = 2} {1\over z^2 + z + 1}\dz &= \int {1\over (z-\zeta_3) (z- \bar{\zeta_3} )} \dz \\ &= \int_{\DD + \bar{ \zeta_3} } {1/(z-\zeta_3) \over (z- \bar{\zeta_3} )} \dz + \int_{\DD + {\zeta_3} } {1/(z- \bar{ \zeta_3} ) \over (z- {\zeta_3} )} \dz \\ &= 2\pi i \qty{1\over z-\zeta_3}\evalfrom_{z=\bar{\zeta_3}} + 2\pi i \qty{1\over z-\bar{\zeta_3}}\evalfrom_{z=\zeta_3} \\ &= 2\pi i \qty{ {1\over \bar{\zeta_3} - \zeta_3 } + {1\over \zeta_3 - \bar{\zeta_3}} }\\ &= 2\pi i \qty{ {1\over \bar{\zeta_3} - \zeta_3 } - {1\over \bar{\zeta_3} - {\zeta_3}} } \\ &= 0 ,\] where \[ \zeta_3=e^{2\pi i/3}=-{1\over2}+i{\sqrt3\over2}, \qquad \zeta_3+\bar\zeta_3=-1, \qquad \zeta_3\bar\zeta_3=1, \] so \(z^2+z+1=(z-\zeta_3)(z-\bar\zeta_3)\).