Contour Integral of $e^z dz$ from $z=1$ to $z=-1$

Question

Evaluate the contour integral of $e^z dz$ along the upper half of the circle absolute value of $z=1$, from $z=1$ to $z=-1$.

I did integral of $e^z dz$ from $z=1$ to $z=-1$ and got $e-e^{-1}$. But the answer is $e^{-1}-e$, can someone explain why?

Answer 1

Note that $e^z$ is analytic and thus the integral $\int_C e^{z}dz$ depends only on the endpoints of $C$.

Given that $\frac{de^z}{dz}=e^z$, if $C$ begins at $(1,0)$ and ends at $(-1,0)$, then we have

$$\int_{1}^{-1}e^z dz=\left.\left( e^z\right)\right|_{z=1}^{z=-1}=e^{-1}-e^{1}$$

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Now, we shall evaluate the integral on the stated contour. Let $z=e^{i\phi}$, $dz=ie^{i\phi}d\phi$, and $\phi$ goes from $0$ to $\pi$. Then,

$$\begin{align} \int_C e^z dz&=\int_{0}^{\pi} e^{e^{i\phi}}ie^{i\phi}d\phi\\\\ &=\left. e^{e^{i\phi}}\right|_{0}^{\pi}\\\\ &=e^{e^{i\pi}}-e^{e^{0}}\\\\ &=e^{-1}-e \end{align}$$

as expected.