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Let $S$ be a subset of $C,$ let $f:S\to C$ be a complex-valued function, and let $c$ be a point of $S.$ Then $f$ is said to be expandable in a Taylor series around c with radius of convergence $r$ if there exists an $r>0$ such that ${B}_{r}\left(c\right)\subseteq S,$ and $f\left(z\right)$ is given by the formula
for all $z\in {B}_{r}\left(c\right).$
Let $S$ be a subset of $R,$ let $f:S\to R$ be a real-valued function on $S,$ and let $c$ be a point of $S.$ Then $f$ is said to be expandable in a Taylor series around c with radius of convergence $r$ if there exists an $r>0$ such that the interval $(c-r,c+r)\subseteq S,$ and $f\left(x\right)$ is given by the formula
for all $x\in (c-r,c+r).$
Suppose $S$ is an open subset of $C.$ A function $f:S\to C$ is called analytic on S if it is expandable in a Taylor series around every point $c$ of $S.$
Suppose $S$ is an open subset of $R.$ A function $f:S\to C$ is called real analytic on S if it is expandable in a Taylor series around every point $c$ of $S.$
Suppose $S$ is a subset of $C,$ that $f:S\to C$ is a complex-valued function and that $c$ belongs to $S.$ Assume that $f$ is expandable in a Taylor series around $c$ with radius of convergence $r.$ Then $f$ is continuous at each $z\in {B}_{r}\left(c\right).$
Suppose $S$ is a subset of $R,$ that $f:S\to R$ is a real-valued function and that $c$ belongs to $S.$ Assume that $f$ is expandable in a Taylor series around $c$ with radius of convergence $r.$ Then $f$ is continuous at each $x\in (c-r,c+r).$
If we let $g$ be the power series function given by $g\left(z\right)=\sum {a}_{n}{z}^{n},$ and $T$ be the function defined by $T\left(z\right)=z-c,$ then $f\left(z\right)=g\left(T\right(z\left)\right),$ and this theorem is a consequence of [link] and [link] .
Prove that $f\left(z\right)=1/z$ is analytic on its domain.
HINT: Use $r=\left|c\right|,$ and then use the infinite geometric series.
State and prove an Identity Theorem, analogous to [link] , for functions that are expandable in a Taylor series around a point $c.$
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