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Show that $\oint log z dz = 2 \pi i$,Where C is the unit circle in the Z plane.

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written 3.8 years ago by |

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

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

Therefore $I =∫_0^{2π}log e^{iθ}. ie^{iθ} \ dθ = i ∫_0^{2π}iθ. e^{iθ} \ dθ $

= $ - ∫_0^{2π}θ. e^{iθ} \ dθ $

= $- [ \frac{θ.e^{iθ}}{i}- \frac{e^{iθ}}{(-1)}]_0^{2π} \,\,\,\,\,\,$ ( using generalized form of integration)

= $- [(-i2πe^{2πi}+e^{2πi} )-(0+1)] =2πi \,\,\,\,\,\,$ [since $e^{2πi}=1$]

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