5

Let

$$z=x+i,w=y+i,$$

Which $x,y\in \mathbb{N}$. We know that

$$\mathrm{Re}(zw)=\sqrt{(z-w{)}^2+2025}$$

We can immediately see that

$$zw=(x+i)(y+i)=xy+i(x+y)-1$$

And

$$\mathrm{Re}(zw)=xy-1$$

We can also see that

$$(z-w)=((x+i)-(y+i){)}^2=(x-y{)}^2$$

Then we have

$$xy-1=\sqrt{(x-y{)}^2+2025}$$

Continue simplify

$$\begin{array}{rl}(xy-1{)}^2& =(x-y{)}^2+2025\\ x^2{y}^2-2xy+1& =x^2-2xy+y^2+2025\\ x^2{y}^2-x^2-y^2+1& =2025\\ (x^2-1)(y^2-1)& =2025\end{array}$$

We know that

$$2025=3^4\times 5^2$$

But we should make sure the factors are in form of $a^2-1$ which $a$ is a positive integer
So $x^2-1$ and $y^2-1$ must be $3$ or $675$
Then we have 2 solutions

$$\{\begin{array}{rl}& x=2\\ & y=26\end{array}\text{or}\{\begin{array}{rl}& x=26\\ & y=2\end{array}$$

Back to $z$ and $w$, we have

$$\{\begin{array}{rl}& z=2+i\\ & w=26+i\end{array}\text{or}\{\begin{array}{rl}& z=26+i\\ & w=2+i\end{array}$$