Calculate the conjugate harmonic function in the indicated domains.
[math]u(x,y)=x^2-y^2+x\,\,\,\,\,\,\,\,0\leq |z|< \infty[/math]
I did this.
[math]u_x=2x+1, \,\,u_y=-2y,\,\, u_{xx}=2,\,\,u_{yy}=-2[/math]
[math]u_{xx}+u_{yy}=0\to 2-2=0[/math]
[math]2x+1=v_y, \,\, -2y=-v_x[/math]
[math]v=\int{(2x+1)dy}\to v=2xy+y+g(x)[/math]
[math]v_x=2y+g'(x)=2y\therefore g'(x)=2y-2y\to g(x)=\int(0)dx=C[/math]
[math]v(x,y)=2xy+y+C[/math]
[math]f(z)=u(x,y)+iv(x,y)[/math][math]f(z)=(x^2-y^2+x)+i(2xy+y+C)[/math]
How do I use the domains?
please help :c
[math]u(x,y)=x^2-y^2+x\,\,\,\,\,\,\,\,0\leq |z|< \infty[/math]
I did this.
[math]u_x=2x+1, \,\,u_y=-2y,\,\, u_{xx}=2,\,\,u_{yy}=-2[/math]
[math]u_{xx}+u_{yy}=0\to 2-2=0[/math]
[math]2x+1=v_y, \,\, -2y=-v_x[/math]
[math]v=\int{(2x+1)dy}\to v=2xy+y+g(x)[/math]
[math]v_x=2y+g'(x)=2y\therefore g'(x)=2y-2y\to g(x)=\int(0)dx=C[/math]
[math]v(x,y)=2xy+y+C[/math]
[math]f(z)=u(x,y)+iv(x,y)[/math][math]f(z)=(x^2-y^2+x)+i(2xy+y+C)[/math]
How do I use the domains?
please help :c