So I'm trying to evaluate the integral $$\oint_{|z|=3} \frac{e^{iz}}{z+i} dz$$ using the Cauchy Integral Formula.
But clearly, the point at $z=-i$ is a singular point, which exists inside our circle $|z|=3$. And this is where I'm confused.
Certainly, the singularity point at $-i$ means $f(z)$ is not analytic everywhere inside our curve since it is undefined at $-i$ .
But according to my textbook, we can apply the Cauchy Integral Formula. But how? The function is not analytic everywhere interior, but this is a condition to use the Cauchy formula. So why is the textbook saying that I can use the Cauchy formula? What am I missing here?
Recall that the Cauchy's Integral Formula says: $$\oint_{\gamma}\frac {f(z)}{z-z_0}dz=2\pi i \cdot f(z_0),$$ where $f(z)$ is analytic in an open set containing the curve $\gamma$ ( taken counter-clockwise) and its interior, and $z_0$ is a point inside $\gamma$.
So in your case we can take $f(z)=e^{iz}$ which in analytic in $\mathbb{C}$, and $z_0=-i$ which is inside the circle $|z|=3$.