Compute $2i\Re(z)\Im(z) = \bar{z} + 3 + i$

Question

Compute: $2i\Re(z)\Im(z) = \bar{z} + 3 + i$

How do you solve this? What do you change $\Re(z), \Im(z)$ to? I tried it like this, but I'm unsure if I've done it right;

$2i·a·bi = a-bi+3+i$

$i(b-1) - 2ab-a-3=0,$ therefore: $b=1$ and $a=-1$, the solution: $z = 1+i$

Is this the correct process?

Answer 1

$$2*i*a*b=a-bi+3+i$$ $$2iab=a+3+i(1-b)$$

How would you find a and b now?

Answer 2

$2iab = 3+a + (1-b)i \Rightarrow a = -3$ and $b = -\frac{1}{5}$.