{{Proofread}} Choose circles $C_1$ and $C_3$ centered at $z_0$ and connect them by path $C_2$ such that $z$ is inside $C_1 + C_2 - C_3 - C_2$ as shown below: :390px This curve and its interior are contained in $B'$, so by Cauchy's Integral Formula: {{begin-eqn}} {{eqn | l = \map f z       | r = \dfrac 1 {2 \pi i} \int_{C_1 + C_2 - C_3 - C_2} \dfrac {\map f w} {\paren {w - z} } \rd w       | c =  }} {{eqn | r = \dfrac 1 {2 \pi i} \int_{C_1 - C_3} \dfrac {\map f w} {\paren {w - z} } \rd w       | c =  }} {{eqn | r = \dfrac 1 {2 \pi i} \int_{C_1} \dfrac {\map f w} {\paren {w - z} } \rd w - \dfrac 1 {2 \pi i} \int_{C_3} \dfrac {\map f w} {\paren {w - z} } \rd w       | c =  }} {{end-eqn}} Since $C_1$ contains $z$ we have for all $w$ on $C_1$: :$\dfrac {\cmod {z - z_0} } {\cmod {w - z_0} } < 1$ Therefore: {{begin-eqn}} {{eqn | l = \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z} } \rd w       | r = \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z + z_0 - z_0} } \rd w        | c = adding and subtracting $z_0$ }} {{eqn | r = \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} \paren {1 - \frac {z - z_0} {w - z_0} } } \rd w        | c = Factoring out $w - z_0$ }} {{eqn | r = \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} } \cdot \frac 1 {\paren {1 - \frac {z - z_0} {w - z_0} } } \rd w        | c =  }} {{eqn | r = \frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0} } \cdot \sum_{n \mathop = 0}^\infty \paren {\frac {z - z_0} {w - z_0} }^n \rd w        | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \frac 1 {2 \pi i} \int_{C_1} \sum_{n \mathop = 0}^\infty \frac {\map f w} {\paren {w - z_0}^{n + 1} } \cdot \paren {z - z_0}^n \rd w        | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{C_1} \frac {\map f w} {\paren {w - z_0}^{n + 1} } \rd w} \paren {z - z_0}^n        | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty a_n \paren {z - z_0}^n        | c = letting $a_n$ equal the value in the large parentheses above, for all $n \ge 0$ }} {{end-eqn}} Similarly, since $z$ lies outside $C_3$ we have for all $w$ on $C_3$: :$\dfrac {\cmod {w - z_0} } {\cmod {z - z_0} } < 1$ Therefore: {{begin-eqn}} {{eqn | l = \frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z} } \rd w       | r = \frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z + z_0 - z_0} } \rd w        | c = Adding and subtracting $z_0$ }} {{eqn | r = \frac 1 {2 \pi i} \int_{C_3} -\frac {\map f w} {\paren {z - z_0} \paren {1 - \frac {w - z_0} {z - z_0} } } \rd w        | c = factoring out $z - z_0$ }} {{eqn | r = -\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {z - z_0} } \cdot \frac 1 {\paren {1 - \frac {w - z_0} {z - z_0} } } \rd w        | c =  }} {{eqn | r = -\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {z - z_0} } \cdot \sum_{n \mathop = 0}^\infty \paren {\frac {w - z_0} {z - z_0} }^n \rd w        | c = Sum of Infinite Geometric Sequence }} {{eqn | r = -\sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{C_3} \map f w \paren {w - z_0}^n \rd w} \paren {z - z_0}^ {-n - 1}        | c =  }} {{eqn | r = -\sum_{k \mathop = -1}^ {-\infty} \paren {\frac 1 {2 \pi i} \int_{C_3} \frac {\map f w} {\paren {w - z_0}^{k + 1} } \rd w} \paren {z - z_0}^k       | c = Re-indexing with $k = -n - 1$ }} {{eqn | r = -\sum_{k \mathop = -1}^ {-\infty} a_k \paren {z - z_0}^k       | c = Letting $a_k$ equal the value in the large parentheses above, for all $k < 0$ }} {{end-eqn}} Combining what has been shown above yields: {{begin-eqn}} {{eqn | l = \map f z       | r = \sum_{n \mathop = 0}^\infty a_n \paren {z - z_0}^n + \sum_{n \mathop = -1}^ {-\infty} a_n \paren {z - z_0}^n       | c =  }} {{eqn | r = \sum_{n \mathop = -\infty}^\infty a_n \paren {z - z_0}^n       | c =  }} {{end-eqn}} {{qed}} Category:Complex Analysis 61p8ciws9o2k8hi7ek43p65qxoycq5o