If $f = u + iv$ and holomorphic in $D$ such that all of its values are on the line $au+bv+c=0$, show that $f$ is a null function on $D$.
I was trying to use C-R equation to get to the result but the most that I get was the $-bu+av + k=0$ for some $k$ real, can I assume something by this?
I guess you mean that, for each $x+iy\in D$ you have $$ au(x,y)+bv(x,y)+c=0 $$ where one among $a$ and $b$ is non zero.
Since the above is an identity, you get $$ a\frac{\partial u}{\partial x}(x,y)+ b\frac{\partial v}{\partial x}(x,y) =0 $$ and, similarly, $$ a\frac{\partial u}{\partial y}(x,y)+ b\frac{\partial v}{\partial y}(x,y) =0 $$ By the Cauchy-Riemann equations, this can be rewritten as $$ \begin{cases} a\dfrac{\partial u}{\partial x}(x,y)- b\dfrac{\partial u}{\partial y}(x,y) =0 \\[4px] a\dfrac{\partial u}{\partial y}(x,y)+ b\dfrac{\partial u}{\partial x}(x,y) =0 \end{cases} $$ Multiply the first by $a$ and the second by $b$, then sum, getting $$ (a^2+b^2)\dfrac{\partial u}{\partial x}(x,y)=0 $$ and, similarly, $$ (a^2+b^2)\dfrac{\partial u}{\partial y}(x,y)=0 $$ Therefore $u$ is constant on $D$. By Cauchy-Riemann again, also $v$ is constant.
The function need not be null, however.