**1 Answer**

written 4.3 years ago by | • modified 4.3 years ago |

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

Therefore $z = e^{iθ}$, $z^2=e^{2iθ}$ and $dz = ie^{iθ} \ dθ$ & θ varies from 0 to π (since C is the upper half of the circle) putting above values in the given integral

Therefore, $I =∫_0^π(e^{iθ}-e^{2iθ}). ie^{iθ} \ dθ = i ∫_0^π(e^{2iθ}-e^{3iθ}) \ dθ$

= $ i[ \frac{e^{2iθ}}{2i}-\frac{e^{3iθ}}{3i}]_0^π = [({\frac{e^{2iπ}}{2}-\frac{e^{3iπ}}{3}})-(\frac{1}{2}-\frac{1}{3})]$

= $ \frac{1}{2} + \frac{1}{3} - \frac{1}{2} + \frac{1}{3} = \frac{2}{3}$

For the lower half of the circle θ varies from π to 2π

Therefore, $I =∫_π^{2π}(e^{iθ}- e^{2iθ}). ie^{iθ} \ dθ = i ∫_π^{2π}(e^{2iθ}- e^{3iθ}) dθ$

= $∫_0^1(x-x+ix^2) = [(\frac{e^4iπ}{2}-\frac{e^6πi}{3})-(\frac{e^2iπ}{2}-\frac{e^3πi}{3})]$

= $ \frac{1}{2} - \frac{1}{3} - \frac{1}{2} - \frac{1}{3} $ [using $e^{nπi} = (-1)^n$ ]

= $- \frac{2}{3}$