Yet another complex analysis problem I need help computing an integral.  My motivation is to understand a standard way to build the holomorphic functional calculus for unbounded operators, though the actual question could probably be a homework problem in a complex analysis class.  Anyway, here goes.
Let $\Omega \subseteq \mathbb{C}$ be the complement of the set $\{iy: y \geq 1, y \leq -1\}$.  Let $\Gamma$ be the contour described by the equation $y^2 - x^2 = \frac{1}{4}$, whose "top half" is oriented from right to left and whose "bottom half" is oriented from left to right.  (The exact equation of $\Gamma$ shouldn't be important, just the overall shape and orientation).  Choose the branch of the holomorphic function $(1 + z^2)^{-1/2}$ on $\Omega$ which takes the value $+1$ at $z = 0$.  I want to prove:
$$\frac{1}{(1+z^2)^{1/2}} = \frac{-1}{2 \pi i} \int_\Gamma \frac{1}{(z - w)(1 + w^2)^{1/2}} dw$$
Don't feel obligated to work out all of the details in your answer - I really should be able to do this on my own...
 A: First, show that $f(z)=\frac{1}{(1+z^2)^{1/2}}$ is holomorphic (analytic) in the set $\Omega$. Then apply the Cauchy's integral formula on $f(z)$:
$$f(z)=\frac{1}{2\pi i}\int_\Gamma \frac{f(w)}{w-z}dw$$
A: As in the original question, let $\Gamma$ be the contour described by the equation $y^2-x^2=1/4$ whose upper part is oriented right-to-left and whose lower part is oriented left-to-right.  We will call the region between these two curves the region interior to $\Gamma$.
Let $k > 0$ and let $C_k = \Gamma_k + \Delta_k^+ + \Delta_k^-$ be the closed contour whose components are defined by
$$\begin{align}
    \Gamma_k &= \{x+iy \colon y^2-x^2=1/4 \text{ and } |x| \leq k\}, \\
    \Delta_k^+ &= \left\{k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\
    \Delta_k^- &= \left\{-k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\
\end{align}$$
which is oriented as in the following picture.

For any $k>0$ the function $f(z) = (1+z^2)^{-1/2}$ is analytic interior to $C_k$, so Cauchy's integral formula tells us that, for $z$ interior to $C_k$,
$$\begin{align}
    \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \int_{C_k} \frac{dw}{(w-z)(1 + w^2)^{1/2}} \\
    &= \frac{1}{2 \pi i} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right),
\end{align} \tag{1}$$
where $g(w,z) = (w-z)^{-1} (1+w^2)^{-1/2}$.  Note that for any $z$ interior to $\Gamma$, $z$ will be interior to $C_k$ when $k$ is large enough.
We will show that the integral over $\Delta_k^+$ goes to $0$ as $k \to \infty$.  The calculation for the integral over $\Delta_k^-$ is identical.
Indeed, we have
$$\begin{align}
    \left|\int_{\Delta_k^+}g(w,z)\,dw\right| &\leq L(\Delta_k^+) \cdot \sup_{w \in \Delta_k^+} |g(w,z)| \\
    &\leq 2 \sqrt{k^2+1/4} \cdot \frac{1}{(k-\text{Re}(z))\sqrt{1+k^2}} \\
    &= O\left(\frac{1}{k}\right).
\end{align}$$
Since $(1)$ holds for any $z$ interior to $\Gamma$ when $k$ is large enough, we can let $k \to \infty$.  Because $\lim_{k \to \infty} C_k = \Gamma$, we get
$$\begin{align}
    \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \cdot \lim_{k \to \infty} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right) \\
    &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{dw}{(w-z)(1 + w^2)^{1/2}}
\end{align}$$
for all $z$ interior to $\Gamma$.
