Let $f$ be an entire function, and $a \in \mathbb{C}$.
Prove that order of $f(z+a)$ is equal to the order of $f(z)$.
My attempt:
Denote $g(z)=f(z+a)$. I tried to take bounds for $f$ and $g$ in the form of $$\Large e^{r^{\text{order} \pm \epsilon}}$$ over circles around $z=0$ that are big enough to cover smaller circles around $z=a$ (the definitions of order all go over circles aroung the origin). But I'm stuck this way, as the radiuses are different, and comparing the values of $f$ and $g$ on circles with different radiuses doesn't imply different or equal order.