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Evaluate (¯z+2z)dz along the circle x2+y2=1.
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x2+y2=1 is a circle with center (0,0) and radius =1  put z=reiθ=1eiθ

∴dz = eiθ . idθ

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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

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