Answer to Question #142502 in Complex Analysis for Doll

Question #142502
Find the original function without finding the corresponding conjugate
u(x,y)=x/x^2+y^2
1
Expert's answer
2020-11-05T16:05:28-0500

For find tho original function we use the Cauchy-Riemann condition:

"\\frac{\\partial u}{\\partial x} = \\frac{\\partial v}{\\partial y}\\\\\n\\frac{\\partial u}{\\partial y} = -\\frac{\\partial v}{\\partial x} \\\\"


"u = \\frac{x}{x^2 + y^2}"

"\\frac{\\partial u}{\\partial x} = \\frac{y^2 - x^2}{(x^2+y^2)^2}=\\frac{\\partial v}{\\partial y}\\\\\n\\frac{\\partial u}{\\partial y} = -\\frac{2yx}{(x^2+y^2)^2} =-\\frac{\\partial v}{\\partial x}"


"v = \\int\\frac{y^2 - x^2}{(x^2+y^2)^2}dy + \\phi(x) = -\\frac{y}{x^2+y^2} + \\phi(x)"

"\\frac{\\partial v}{\\partial x} = \\frac{2yx}{(x^2+y^2)^2} + \\phi\\prime(x) = \\frac{2yx}{(x^2+y^2)^2}"

"\\phi\\prime(x) = 0 ,\\phi(x) = C"

So: "v(x,y) = -\\frac{y}{x^2+y^2} + C" , and "w(u,v) = \\frac{x}{x^2+y^2} - i\\frac{y}{x^2+y^2}"


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