Prove complex integral equality

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 $$2 \pi i ( z -5)$$.

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?

• Why does it converge absolutely for the unit circle?
– Pegi
Jan 22 at 15:12
• Because it converges in an open set containing the unit circle, and therefore on a circle of slightly smaller radius about $0$. Jan 22 at 17:02

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!