My question is with reference to wireless communication.
There are $N$ independent and identically distributed random channels i.e., $Z_i \in \{ Z_1, Z_2,\ldots, Z_N\}$ with PDFs $f_{Z_i}(z)$ ordered in ascending order. Then using order statistics the PDF of $Z_1 = \text{min}_i(Z_i)$ is given by
$$f_{Z_1}(z) = N (1-F_{Z_i}(z))^{N-1}f_{Z_i}(z)\tag{1}$$
Will equation (1) will be valid if the random channels are not independent and not identically distributed?
No, it will not be valid. Think about where the exponent comes from. Also, if you have different distributions, how can there be only an arbitrarily picked $f_{Z_i}$ in your (1)? That doesn't make sense. You should probably build up a little more intuition here!
If you're working on ordered statistics, calculating the PDF of $Z_\min=\min(\{Z_i\})$ for i.i.d. RVs is probably something you want to do at least once yourself to get yourself warm with the common tricks when dealing with collections of random variables.
The derivation of $f_{Z_\min}$ is actually pretty straightforward. I'll give you a start!
\begin{align} f_{Z_\min}(z) &=\frac{\mathrm d}{\mathrm d\,z}F_{Z_\min}(z) \tag2\\ &=\frac{\mathrm d}{\mathrm d\,z}P({Z_\min}\le z)\tag3\\% &=\frac{\mathrm d}{\mathrm d\,z}P(Z_1\le z \,\wedge\, Z_2\le z \,\wedge\,Z_3\le z \,\wedge\,\ldots\,\wedge\,Z_N\le z )\tag4\\% \end{align}
Now, at this point your calculation starts depending on whether $Z_i$ are independent! Try with independent variables $Z_i, \quad i=1,\ldots,N$ first. Remember the definition of what it means for random variables to be independent. Then, don't forget to apply the chain rule of derivatives.
If all you want is a yes or no answer -- no.
The proof is easy: in the extreme case, let each $Z_i = Z$. Then the pdf of the minimum is just the pdf of $Z$, and that doesn't equal your (1).
