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Evaluate ∫(¯z+2z)dz along the circle x2+y2=1.
1 Answer
written 8.7 years ago by | • modified 8.7 years ago |
x2+y2=1 is a circle with center (0,0) and radius =1 put z=reiθ=1eiθ
∴dz = eiθ . idθ
and, = e^{-iθ}
∴ = . ie^{iθ}dθ
= i . dθ
= i [1θ +]_0^{2π}
= i ]
Now, = \cos 4π + i \sin 4π = 1 + i(0) = 1
∴ = i{ 2π + - }
∴ = 2iπ