Prove complex integral equality

Question

Suppose $\triangle$ is the open unit disk and $\overline{\triangle}$ be it’s closure (closed unit disk). Let $f$ be holomorphic in an open set containing the set $D = \mathbb{C} - \overline{\triangle}$ such that $ lim_{z \to \infty} f(z) = 5$. Show for $z \in D$, $$\frac{1}{2 \pi i} \int_{\partial \triangle} \frac{f(\zeta)}{\zeta -z} d\zeta = 5 - f(z) $$

I tried to use a proof similar to that of Cauchy’s formula using the key hole contour with the smaller circle being the integral above and the radius of the bigger circle going to infinity. However, I don’t know how to show that the contour integral of the bigger circle is $ 2 \pi i ( z -5)$.

Answer 1

Hint: $f$ has a Laurent expansion $f(\zeta) = 5 + a_1/\zeta+ a_2/\zeta^2 + \ldots$ which converges absolutely for $|\zeta| = 1$. What do you get for your integral with $f(\zeta)$ replaced by each of these terms?

Answer 2

Since $f$ is analytic on $\mathbb{C}\setminus \bar\Delta$, we assert that for $R>|z|$

$$\begin{align} \oint_{\partial \Delta}\frac{f(\zeta)}{\zeta-z}\,d\zeta&=\oint_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta -2\pi i f(z)\\\\ &=\int_0^{2\pi}\frac{f(Re^{i\phi})}{Re^{i\phi}-z}\,iRe^{i\phi}\,d\phi-2\pi i f(z) \end{align}$$

Letting $R\to \infty$ we find that

$$\oint_{\partial \Delta}\frac{f(\zeta)}{\zeta-z}\,d\zeta=2\pi i (5-f(z))$$

as was to be shown!