How should I tackle this integral? I can't use the fact that $\sigma$ is a closed curve because the function is not defined in z=2 which is inside the curve. Any tip on how to proceed is welcomed.
2026-03-31 17:49:31.1774979371
Integral $\int_{\sigma}{{e^zdz}\over{z-2}}$ with $\sigma:[0,2\pi] \to C, \sigma (t)=4e^{it}$
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Using cauchy's integral theorem $e^2= {1\over{2\pi i}}\int_{\sigma}{e^zdz\over{z-2}} $
Then $\int_{\sigma}{e^zdz\over{z-2}}=e^22\pi i $