Zeros satisfy π/n < 1 precisely for n ≥ 4. Thus 14 zeros, n = 4,…,17, lie inside the unit circle; none lies on it. The curve $e^{2it}$ traverses that circle twice counterclockwise. The argument principle gives integral $2\pi i(2)(14)=56\pi i$, so α = 56.
More Complex Integration and Residues questions
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