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.


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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