Question #313819

Using the Cauchy –Riemann equations verify the following is analytic or not


i) π‘₯^2 βˆ’ 𝑦^2 + 2𝑖π‘₯𝑦


ii) π‘₯^2 + 𝑦^2 βˆ’ 2𝑖π‘₯𝑦

1
Expert's answer
2022-03-19T02:40:17-0400

i:u(x,y)=x2βˆ’y2v(x,y)=2xyβˆ‚uβˆ‚x=2x,βˆ‚uβˆ‚y=βˆ’2yβˆ‚vβˆ‚x=2y,βˆ‚vβˆ‚y=2xβˆ‚uβˆ‚x=βˆ‚vβˆ‚y,βˆ‚uβˆ‚y=βˆ’βˆ‚vβˆ‚xβ‡’f  is  analyticii:u(x,y)=x2+y2v(x,y)=βˆ’2xyβˆ‚uβˆ‚x=2xβˆ‚vβˆ‚y=βˆ’2xβˆ‚uβˆ‚xβ‰ βˆ‚vβˆ‚yβ‡’f  is  not  analytici:\\u\left( x,y \right) =x^2-y^2\\v\left( x,y \right) =2xy\\\frac{\partial u}{\partial x}=2x,\frac{\partial u}{\partial y}=-2y\\\frac{\partial v}{\partial x}=2y,\frac{\partial v}{\partial y}=2x\\\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\Rightarrow f\,\,is\,\,analytic\\ii:\\u\left( x,y \right) =x^2+y^2\\v\left( x,y \right) =-2xy\\\frac{\partial u}{\partial x}=2x\\\frac{\partial v}{\partial y}=-2x\\\frac{\partial u}{\partial x}\ne \frac{\partial v}{\partial y}\Rightarrow f\,\,is\,\,not\,\,analytic


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