I was simplifying this problem for a class exercise the other day that looked something like this:
$$(5x+9)(5x-9)$$
Obviously the simplified version of this is $25x^2-81$, but I wondered to myself, what if I replaced $x$, with the imaginary unit, $i$. So I did, and this is what I got. Am I correct?
$$\begin{align}&(5i+9)(5i-9)\\&=25i^2-81\\&=25(-1)-81\\&=-25-81\\&=-106\end{align}$$
On
I wouldn't say $25x^2 - 81$ is a solution to $\left({5x + 9}\right)\left({5x - 9}\right)$. Those aren't equations, they're just formulas; there's nothing to solve. Though I suppose if the point of the exercise is to expand the product, you've found the solution to the exercise. But the formula itself doesn't have anything to solve.
What you have done is found the value of the function defined by $25x^2 - 81$ evaluated at $x = i$. You have also defined the function to have imaginary numbers in its domain, which is fine. If the creator of the function defined the domain to be only real numbers, the function would be undefined at $x = i$. Many books don't specify the domain of functions, leaving the reader to guess what is implied (which in my opinion is irresponsible).
Yes, this works; the fact that $(5x+9)(5x-9)=25x^2-81$ can be interpreted as a more general consequence of the "algebraic" properties of multiplication and addition. In particular, both the real and complex numbers form a field, which is a structure that allows for addition and multiplication, specifying that we get properties like: $$ab=ba$$ $$a(b+c)=ab+ac$$ $$a+b=b+a$$ and some more - but the above are sufficient to prove that $(5x+9)(5x-9)=25x^2-81$ - therefore, since both real and complex multiplication and addition satisfy the above, it will hold. So, replacing $x$ with $i$ must still be true.