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  "body": "\n## Definition\n\nLet $A$ be a complex [[Banach algebra|9KYWN94QDN]] with unit $e$, and let $a \\in A$ with [[spectrum|EH0CZ19RKP]] $\\sigma(a)$.\nLet $U \\subseteq \\C$ be an open set with $\\sigma(a) \\subseteq U$, and let $f \\colon U \\to \\C$ be [[holomorphic|0NAHXR1FRH]].\nLet $\\Gamma$ be a finite union of closed piecewise smooth curves in $U \\setminus \\sigma(a)$.\nAssume that the winding number of $\\Gamma$ is $1$ around each point of $\\sigma(a)$ and $0$ around each point of $\\C \\setminus U$.\nThe element $f(a) \\in A$ is the [[contour integral|06ZGPYTWK8]]\n$$f(a) = \\frac{1}{2\\pi i} \\int_\\Gamma f(\\lambda) \\, (\\lambda e - a)^{-1} \\, d\\lambda .$$\nThe map $f \\mapsto f(a)$, from the algebra of holomorphic functions on $U$ to $A$, is the *holomorphic functional calculus* of $a$.\n\n## Remarks\n\nThe integrand is continuous, since $\\lambda \\mapsto (\\lambda e - a)^{-1}$, the [[resolvent|J5YNNKSB9P]], is defined and holomorphic off $\\sigma(a)$.\nBy Cauchy's theorem for $A$-valued holomorphic functions, $f(a)$ does not depend on the choice of $\\Gamma$.\nExample: for $f(\\lambda) = \\lambda^n$ and $\\Gamma$ a large circle $\\abs{\\lambda} = R$ with $R \u003e \\norm{a}$, expand $(\\lambda e - a)^{-1} = \\sum_{k \\ge 0} \\lambda^{-k-1} a^k$; only the term $k = n$ survives the integral, so $f(a) = a^n$.\nNon-example: $f(\\lambda) = \\bar\\lambda$ is not holomorphic, so the definition does not apply; for non-normal operators there is no reasonable way to define $\\bar a$ by this method.\nThe calculus is an algebra homomorphism, it is continuous for locally uniform convergence of $f$, and $\\sigma(f(a)) = f(\\sigma(a))$.\nFor a normal operator on a Hilbert space it agrees with the [[continuous functional calculus|JB7S6WPZ77]] on holomorphic functions.\nWhen $f$ is $1$ near one closed and open part of $\\sigma(a)$ and $0$ near the rest, $f(a)$ is the [[Riesz projector|1AR19ZQQM2]] of that part.\n",
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