$\oint_{\gamma}\frac {\cos z}{z}dz$ on the curve $\gamma = \gamma_1\colon|z| =1$ and $\gamma = \gamma_2\colon|z|=3$. I calculated on $|z| = 1$ but I don't see why it would be different if I would calculate on $|z| = 3$, since I apply the residue formula at $z_0 = 0$. This is what I did:
$$\oint_{\gamma}\frac {\cos z}{z}dz = 2\pi i \operatorname{Res}(f,0)$$ from the Residue Theorem.
We have $\operatorname{Res}(f,0) = \lim_{z\to0}z\frac {\cos(z)}{z} = 1.$ Thus the integral is $I = 2\pi i$ and because the only singular point which is a simple point is contained in both curves then the answer must be the same right?
Alternatively, you can use Cauchy's integral formula $$f^{(n)}(a)=\frac{n!}{2\pi i}\int\limits_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz \tag{1}$$ In this case, since $0$ is "inside" $|z|=1$ and $|z|=3$ $$f(0)=\frac{1}{2\pi i}\int\limits_{\color{red}{|z|=1}}\frac{f(z)}{z}dz$$ $$f(0)=\frac{1}{2\pi i}\int\limits_{\color{red}{|z|=3}}\frac{f(z)}{z}dz$$ where $f(z)=\cos{z}$ and $f(0)=\cos{(0)}=1$, thus $$2\pi i=\int\limits_{\color{red}{|z|=1}}\frac{f(z)}{z}dz=\int\limits_{\color{red}{|z|=3}}\frac{f(z)}{z}dz$$