Suppose that f is a holomorphic function on $G=\{z:Re\;z>1\}$ and that $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ on G. Show that are real constant $\alpha$ and a complex constant $\beta$ such that $f(z)==-i\alpha z+\beta$ on G
My idea:
since $f(z)$ is holomorphic then $u_x=v_y$
since $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\rightarrow \frac{\partial u}{\partial x}=-\frac{\partial v}{\partial y}\\ v_y=0\rightarrow v=a$
but i dont get answer can any one help me
From Cauchy-Riemann equations, we have $u_x=v_y$ and $u_y=-v_x$ and using the given condition $u_x+v_y=0$, we have $u=\phi(y)$ and $v=\psi(x)$.
Now, $u_y=-v_x$ implies $\phi^\prime (y)=-\psi^\prime (x)$. The LHS is a function of $y$ and the RHS is a function of $x$ and they can be identical only if they both equal a constant (say $K$) thus giving $u=\phi(y)=Ky+C_1$ and $v=\psi(x)=-Kx+C_2$ for some constants $C_1,C_2$ thus giving, $$\small f=u+iv=Ky+C_1+i(-Kx+C_2)=K(y-ix)+(C_1+iC_2)=-Ki(x+iy)+(C_1+iC_2)=\ldots$$