Suppose $f : \mathbb{C} \to\mathbb{ C} $ is a holomorphic function such that the real part of $f''(z)$ is strictly positive for all $ z \in \mathbb{C}$. What is the maximum possible number of solutions of the equation $ f (z) = az + b$, as $a $ and $b$ vary over complex numbers?
I have tried a lot but still confused.
The part you may be missing is that if $f$ is entire then $f''$ is entire as well. You are told that it has positive real part, but the only entire functions with positive real part are constants.
So $f''(z)=c$ for all $z$, for some fixed $c \in \mathbb{C}$. Therefore $f'(z)$ is... and therefore $f(z)$ itself is...
EDIT: by popular demand, let's proceed. Since $f''$ is constant, $f'$ is linear and $f(z)=Az^2+Bz+C$ is a polynomial of degree $2$ ($A$ must be positive, so the degree is not less than that). The equation $f(z)=az+b$ therefore has exactly two solutions, counted with multiplicity. You can also say that it has at most two solutions.
Indeed, you can take $f(z)=z^2$ (which has $f''(z)=2$, satisfying the requirement of positive real part), and $a=0$, $b=1$. The equation is now $z^2=1$ which has exactly two distinct solutions.