Getting rid of complex zeros of function with zeros the primes?

Question

From our Note: simple real function with zeros greater than one the primes

simple real function with zeros greater than one the primes: $j_1(x)=(\sin(\pi x))^2+(\sin(\pi \frac{\Gamma(x)+1}{x}))^2$. For $x>0$ the function is continuous.

$j_1$ has complex zeros for $\Re(x)>1$.

Q1 Can we remove the complex zeros while keeping the primes as zeros? Preferably we want to keep continuity.

Trying one approach. Define $f_1(x)=\sin(\pi x)$ and $f_2(x)=\sin(\pi \frac{\Gamma(x)+1}{x})$

We have $j_1(x)=f_1(x)^2+f_2(x)^2$. The complex zeros of $j_1$ which are not integers are $f_1(\rho)=\pm i \cdot f_2(\rho)$.

Assume $s$ is complex number, not an integer and $\Re(s)>1$.

Let $g(s)$ be some complex function, possibly constant.

Define $j_2(s,g(s))=g(s) f_1(s)^2 + f_2(s)^2$.

$j_2$ keeps the zeros at primes and introduces zeros

$$\sqrt{g(\rho)}=\frac{f_2(\rho)}{f_1(\rho)} \qquad (1)$$

If (1) doesn't have solution, we are done. If it has solutions for all $g$, then $\frac{f_2(\rho)}{f_1(\rho)}$ will be very close to surjective.

Another approach might be to take $F(f_1,f_2)=0$.

Observe that we allow zeros with $\Re(s) \le 1$.

Working numerically is very hard and introduces significant errors even for $s=41$.

Plot over the reals:

Here is X-Ray,$\Re(j_1(z))=0$ is plotted red and $\Im(j_1(z))=0$ is plotted blue. The intersections of the curves are zeros.

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