Since the closed curve is the circle we use polar coordinates to evaluate above integral i.e. $z = re^{iθ}$ , since r =1

Therefore $z = e^{iθ}$ & $ \bar{z} =e^{-iθ}$ ∴ $dz = ie^{iθ} \ dθ$ & θ varies from 0 to 2π putting above values in the given integral

Therefore $I =∫_0^{2π}(e^{-iθ}+2 e^{iθ}). ie^{iθ} \ dθ = i ∫_0^{2π} (1+2e^{2iθ}) \ dθ $

= $ i[θ+ \frac{2 e^{2iθ}}{2i}]_0^{2π}$

= $ i [(2π-ie^{4πi} )-(0-i)]$

= $ 2πi +e^{4πi} -1 = 2πi $ (since $e^{4πi}=1 ;\ \frac{1}{i} = -i$ )