# Let $f=u+iv$, show that $f(z)=-icz+d$ with $c\in \mathbb{R}, d\in\mathbb{C}$

Let $f$ be analytic in $\{z\in \mathbb{C} | \Re(z)>1\}$ such that $f=u+iv$ and $\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0$ in $U$. Show that there exist $c \in \mathbb{R}, d /in \mathbb{C}$ such that $f(z)= -icz +d$.

So if $f$ is analytic it satisfies Cauchy Riemann equations and there fore the partials equality would imply that either $u$ or $v$ are constant. I asked my teacher and he said I should integrate but I'm lost, anything I'm missing?

Thanks

According to Cauchy Riemann equations we have$$\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\\\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$$which by considering $\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=0$ leads to $$\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}=-\dfrac{\partial v}{\partial y}=0$$since both $u$ and $v$ are real we have that$$u=f_1(y)\\v=f_2(x)$$and by substitution in $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$ we get $$f_1^{'}(y)=f_2^{'}(x)$$and both must be real constants (why?) so we have$$f_1^{'}(y)=f_2^{'}(x)=C\qquad,\qquad C\in\Bbb R$$or$$f_1(y)=Cy+d_1\\f_1(y)=Cx+d_2$$and by integration we obtain$$f(x,y)=u+iv=Cy+d_1-i(Cx+d_2)=-iCz+D$$where $C\in\Bbb R$ and $D\in\Bbb C$