if $\int_{\gamma}f(z)dz=0$, then f is analytic at all points within $\gamma$
I consider it to be false since if I consider $f(z)=\frac{1}{z^2}$, and $\gamma:=|z|=1$, by Cauchy's integral formula, is satisfied $$\int_{\gamma}\frac{1}{z^2}dz=0$$ However $f(z)=\frac{1}{z^2}$ is analytic in $\mathbb{C}-\lbrace 0\rbrace$, that is, the function f is not analytic at all points of $\gamma$.
therefore the result is false.
Find a result of Morera's theorem, which adds the continuity hypothesis, on the contour, which guarantees that the previous result is true.
Find $\displaystyle\int_{\gamma}\frac{\sin{z}}{(z-\zeta)^2}dz$ where $\gamma:=|z|=1$ and $|\zeta|<1$
By Cauchy's integral formula if $\zeta\in int \gamma$, it must be fulfilled that $\zeta=0$, now $$\displaystyle\int_{\gamma}\frac{\sin{z}}{(z-\zeta)^2}dz=\frac{2\pi i f^{'}(\zeta)}{1!}=2\pi i\cos{\zeta}$$
I have a doubt regarding, if this is still fulfilled this equation for $|\zeta|>1$, I think that it is not fulfilled since the singularities of $f (z)$ would no longer be in $int\gamma$
If $f$ is analytic within $\gamma$, except in $z_{0}$ and $(z-z_0)f(z)$ is analytic inside $\gamma$, then $\displaystyle\int_{\gamma}f(z)dz\neq 0$
I do not understand this last section, I must verify if it is true false, any suggestion is gratefully received.
In the same way, I would like to know if the other sections are well done, I appreciate your understanding
You can prove this using the Residue Theorem. The value of the integral is $2\pi i \lim (z-z_0)f(z)$. If the integral is $0$ then this limit must be $0$. Now define $g(z)=(z-z_0) f(z)$ if $z \neq z_0$ and $0$ if $z=z_0$. Then $g$ is analytic and it has a zero at $z_0$. So we can write $g(z)=(z-z_0)h(z)$ where $h$ is analytic . But this makes $f$ anayltic at $z_0$ Hence, as long as $f$ is not analytic at $z_0$ the integral cannot vanish.