Let $f(s)$ be a general Dirichlet series
$$f(s)=\sum_{n=1}^\infty a(n)e^{-s\lambda(n)}$$
where $\lambda(n)$ and $a(n)$ are given by
It is easy to prove that the Bohr function associated with this Dirichlet series is
$$F(z_1,z_2,z_3)=\cos(iz_1)-\frac{1}{2}i\sin(iz_2)(1+\cos(iz_1))+\frac{1-2e^{-z_3}}{2-e^{-z_3}}$$
Now I was told to find the set of values taken by $F$ at any $s$ with a real part $0$:
$$U_f(0)=\{F(Z):\Re(Z)=0\}$$
I tried to plot some values of the function using python and got the following result.
It seems to me that this is a triangle with rounded corners. So I claim (maybe falsely) that for some $u\in U_f(0)$ we could always find $0\leqslant\theta_1,\theta_2,\theta_3$ such that $\theta_1+\theta_2+\theta_3=1$ and there exists a point $v=(-1)\theta_1+(1-i)\theta_2+(1+i)\theta_3$ (these three points are said to be significant in the hint) such that
$$|u-v|\leqslant 1$$
Therefore I expanded $u$ and got
$$F(iy_1,iy_2,iy_3)=\left(\cos y_1+1+\frac{\cos y_3+1}{4\cos y_3-5}\right)+i\left(\frac{\cos y_1+1}{2}\sin y_2-\frac{3\sin y_3}{4\cos y_3-5}\right)$$
However, I couldn't do much after this. I think it is possible to prove that this set is closed (might be wrong) and therefore
$$U_f(0)=\overline{U_f(0)}=\overline{V_f(0)}$$
which is the set of values taken by $f(it)$. Then
$$\begin{align*}
f(it)
={}&\frac{-3\cos(t\log 3)}{4}+\cos(t)+\frac{1}{2}\\
&+i\left(\frac{\sin(t\log 2-t)}{4}+\frac{\sin(t\log 2+t)}{4}+\frac{3\sin(t\log 3)}{4}+\frac{\sin(t\log 2)}{2}\right)\\
&+\frac{-3^{1-it}}{4(2\cdot 3^{it}-1)}
\end{align*}$$
Even though plotting this function gives the same thing, I believe this is even worse than bashing on $U_f(0)$.
This problem is from Apostol's Modular Functions and Dirichlet Series in Number Theory. I tried to bash out the answer as you can see. But I believe there should be some better ideas and some smarter ways to tackle this problem. Any help will be appreciated; thanks in advance.