Using this theorem I am supposed to solve 4.35 here. Carefully look at the theorem.
So first thing I do is let f(u) = 1/(u-4), because f is holomorphic on ℂ\{4} which contains D[0, 3]. But this transforms my integral into this, which has w ≠ w'.
So I pick a different $f(u) = u/(u^2 - 2u + 8)$, which does result in w = w'... but this new f is only holomorphic on ℂ\{-2, 4}, and D[0, 3] ⊈ ℂ\{-2, 4}.
So I am out of ideas and I need to use the given theorem. How? I need to use the Cauchy theorem.
The hint in the previous exercise says to use a partial fraction decomposition. The integrand decomposes as
$$ \frac{1}{z^2-2z-8} = \frac{1}{6(z-2)} - \frac{1}{6(z+4)}. $$
Okay, now let's integrate over the circle of radius $3$ centered at the origin (I'll call this $C$).
\begin{align*} \oint_C \frac{1}{6(z-2)} - \frac{1}{6(z+4)} \ dz &= 2\pi i (1/6) - 0 \\ &= \frac{\pi i}{3} \end{align*}
Why? The first integral is $\frac{\pi i}{3}$, as the singularity at $z = 2$ is inside the circle. The function in the form of the Integral Formula is $f(z) = \frac{1}{6}$, a constant function. For the second integral, the singularity at $z = -4$ is outside the circle, and by the Cauchy Theorem, the integral is zero.
This should be what you need.