Cauchy Integral Formula for z outside disc

Question

If $ f(z) $ is analytic for $ |z| \geq 1 $, and $ f(z) $ is bounded as $ z $ approaches infinity, then how can I derive an equivalent of Cauchy' Integral Formula to express $ f(z) $ for $ z $ outside unit disc?

Answer 1

Try applying an inversion ($ z \to \frac{1}{z} $, which is holomorphic) to re-express this problem in terms of the unit disc and Cauchy's formulas. Since $ f $ is bounded outside the unit disc, $ f \left ( \frac{1}{z} \right ) $ will be bounded (and therefore holomorphic) inside the unit disc.

Answer 2

Since $f(z)$ is analytic for $|z|\ge 1$ and bounded as $|z|\to \infty$, we can represent it by the Laurent expansion

$$f(z)=\sum_{n=0}^\infty a_nz^{-n}$$

for $|z|\ge 1$.

METHODOLOGY $1$: Integration of Laurent Series

Therefore, we have for $|z|>R$ for any $R\ge 1$

$$\begin{align} \oint_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta&=\sum_{n=0}^\infty a_n \oint_{|\zeta|=R} \frac{1}{\zeta^n(\zeta-z)}\,d\zeta \tag 1\\\\ &=\sum_{n=0}^\infty a_n \left(2\pi i \,\text{Res}\left(\frac{1}{\zeta^n(\zeta-z)},\zeta=0\right)\right)\\\\ &=-2\pi i \sum_{n=1}^\infty a_n \left(\frac{1}{z^n}\right)\\\\ &=2\pi i (f(\infty)-f(z)) \end{align}$$

where $f(\infty)=\lim_{z\to \infty}f(z)=a_0$. In $(1)$, the orientation of the contour $|\zeta|=R$ is taken counter-clockwise, which conforms with Cauchy's Integral Formula.

Finally, we can write for $|z|>R\ge 1$

$$\bbox[5px,border:2px solid #C0A000]{f(z)=f(\infty)-\frac1{2\pi i}\oint_{|\zeta|=R\ge 1}\frac{f(\zeta)}{\zeta-z}\,d\zeta} \tag 2$$

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METHODOLOGY $2$: Contour Deformation and Application of the Residue Theorem

Alternatively, using the residue theorem we have for $R'>|z|>R\ge 1$

$$\oint_{|\zeta|=R'}\frac{f(\zeta)}{\zeta-z}\,d\zeta-\oint_{|\zeta|=R}\frac{f(\zeta)}{\zeta-z}\,d\zeta=2\pi i f(z) \tag 3$$

Letting $R'\to \infty$ in the first integral on the left-hand side of $(3)$ reveals that

$$\lim_{R'\to \infty}\oint_{|\zeta|=R'}\frac{f(\zeta)}{\zeta-z}\,d\zeta=2\pi i f(\infty)$$

Since the other terms in $(3)$ are independent of $R'$ we find that

$$\bbox[5px,border:2px solid #C0A000]{f(z)=f(\infty)-\frac{1}{2\pi i}\oint_{|\zeta|=R\ge 1}\frac{f(\zeta)}{\zeta-z}\,d\zeta}$$

which agrees with the result in $(2)$!

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METHODOLOGY $3$: Residue at Infinity

And finally, using the concept of the Residue at Infinity, we can write

$$\begin{align} \oint_{|\zeta|=R\ge 1}\frac{f(\zeta)}{\zeta-z}\,d\zeta &=-2\pi i \text{Res}\left(\frac{f(\zeta)}{\zeta-z},\zeta=z\right)+2\pi i \text{Res}\left(\frac{f(\zeta)}{\zeta-z},\zeta=\infty\right)\\\\ &=-2\pi i f(z)-2\pi i \text{Res}\left(-\frac{1}{w^2}\frac{f(1/w)}{1/w - z},w=0\right)\\\\ &=2\pi i (f(\infty)-f(z)) \end{align}$$

whereupon solving for $f(z)$, we recover $(2)$ as expected.