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This function is defined on $\mathbb{C}-\{-2,2\}$. My problem is whether it is analytic on this set or not. If yes, then the integral is zero since $\gamma$ is a closed curve. If no, I have to calculate anyway. I could not solve the Cauchy Riemann equations. Yet, I know that $\frac{1}{z^2}$ is analytic on $\mathbb{C}- \{0\}$. At this point can I say that, in general,

$f(z)$ is analytic implies $f(z-a)$ is analytic with a proper domain (of course this statement is very intuitive but this is all I have).

I did the following for this statement:

Let $\phi(x,y)=x-4+iy$

Then, $f(z-4)=f(\phi(x,y))=u(x-4,y)+iv(x-4,y)=u(\phi(x,y))+v(\phi(x,y))$

Since $f(x+iy)=u(x,y)+iv(x,y)$ is analytic we have $u_x=v_y$ and $u_y=-v_x$.

Since $\phi_x = 1$ and $\phi_y = 1$, by the chain rule, we have

$u(\phi(x,y))_x = v(\phi(x,y))_y $ and $v(\phi(x,y))_x = -u(\phi(x,y))_y $

Therefore, $f(z-4)$ is also analytic.

Is this a good argument? What are the technical issues I have to consider? Is there a general theorem or idea that I do not know? Can you help me with these issues? Thanks in advance.

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All that matters is whether the function is analytic within the region whose boundary you are integrating over. As $\frac{1}{z^2-4}$ has no points of singularity in the disc $|z|\le 1$, the function is analytic here. Thus, the integral $= 0$

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