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  "body": "\n## Definition\n\nLet $c \\in \\C$ and $R \u003e 0$, and let $D = \\set{z \\in \\C : \\abs{z - c} \u003c R}$.\nFor $\\zeta$ on the circle $\\abs{\\zeta - c} = R$ and $z \\in \\C$ with $z \\ne \\zeta$, the *Herglotz–Riesz kernel* of $D$ is\n$$H(\\zeta, z) = \\frac{(\\zeta - c) + (z - c)}{(\\zeta - c) - (z - c)} = \\frac{\\zeta + z - 2c}{\\zeta - z}.$$\n\n## Remarks\n\nFor fixed $\\zeta$, the function $z \\mapsto H(\\zeta, z)$ is [[holomorphic|0NAHXR1FRH]] on $D$.\nIts real part is the [[Poisson kernel|TCYSWCTYYM]]: with $a = \\zeta - c$ and $b = z - c$,\n$$\\operatorname{Re} H(\\zeta, z) = \\operatorname{Re} \\frac{(a + b)(\\bar a - \\bar b)}{\\abs{a - b}^2} = \\frac{R^2 - \\abs{z - c}^2}{\\abs{\\zeta - z}^2},$$\nbecause $b \\bar a - a \\bar b$ is purely imaginary.\nExample: for the unit disc ($c = 0$, $R = 1$), $\\zeta = 1$ and $z = 1/2$, one gets $H = 3$, and the Poisson kernel $(1 - 1/4)/(1/4)$ is also $3$.\nNon-example: the Cauchy kernel $\\zeta/(\\zeta - z)$ is also holomorphic in $z$, but at $\\zeta = 1$, $z = 1/2$ it equals $2$, so its real part is not the Poisson kernel.\nFor $c = 0$ the two kernels are related by $H(\\zeta, z) = 2\\zeta/(\\zeta - z) - 1$.\nIf $u$ is harmonic on a neighbourhood of $\\overline{D}$, the [[Schwarz integral formula|61WT6CE08W]] reads\n$$u(z) + i v(z) = \\frac{1}{2\\pi} \\int_0^{2\\pi} H(c + R e^{i\\theta}, z) \\, u(c + R e^{i\\theta}) \\, d\\theta,$$\nwhere $v$ is the harmonic conjugate of $u$ on $D$ with $v(c) = 0$.\nThe same kernel appears in the [[Herglotz representation theorem|9ZBAX4ZPX0]].\nOther names in use are Herglotz kernel and Schwarz kernel.\n",
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