Answer to Question #142504 in Complex Analysis for Doll

Question #142504
Find the original function without finding the corresponding conjugate
u(x,y)=e^xcosy
1
Expert's answer
2020-11-08T18:47:15-0500

"\\displaystyle u(x, y) = e^x \\cos(y)\\\\\n\n\\frac{\\partial u}{\\partial x} = e^x \\cos(y) \\\\\n\n\\textsf{A harmonic function satisfies the}\\\\\\textsf{Cauchy Riemann's equation}\\\\\n\n\n\\frac{\\partial v}{\\partial y} = \\frac{\\partial u}{\\partial x} = e^x \\cos(y)\\\\\n\n\n\\therefore v = \\int e^x \\cos(y)\\, \\mathrm{d}y = e^x\\sin(y) + C\\\\\n\n\n\\begin{aligned}\nf(z) &= u(x, y) + iv(x, y) \n\\\\&= e^x(\\cos(y) + i\\sin(y)) + C \n\\\\&= e^x\\cdot e^{iy} + C \\\\&= e^{x + iy} + C = e^z + C\n\\end{aligned}\n\n\\therefore \\textsf{The original function is}\\, f(z) = e^{z} + C\\\\\n\\textsf{where}\\, C\\, \\textsf{is an arbitrary constant.}"


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