# Evaluate the integral $\int _\gamma (e{^z}^{2} + \overline{z}) dz$

First part of the question asks me to state the path integral $\int_\gamma f$, which I defined as: $$\int_\gamma f = \int^b_a f(\gamma(t))\gamma ' (t) dt$$

And the second part asks to evaluate the integral $$\int _\gamma (e{^z}^{2} + \overline{z}) dz$$ where $\gamma$ is the positively oriented unit circle.

Does this mean that $\gamma(t) = e^t + t \ (0 \leq t \leq 1)$ ?

• $\gamma (t) = e^{2 \pi i t} \quad t \in (0,1)$, is a fairly standard parameterization of the unit circle – Triatticus May 27 '15 at 7:52
• I think inside the integral you have $\;e^{z^2}\;$ ? – Timbuc May 27 '15 at 7:58

Since $e^{z^2}$ is analytic on and inside the curve given and since $\overline{z}=1/z$ on this curve, the integral is $0 + 2\pi i$.