Proof of a complex identity This might be a very obvious one, but I am stuck on this from a long time. 
If $F(s) = M(s) + N(s)$ where $M(s)$ is even polynomial function and $N(s)$ is odd polynomial function (where $s$ is a complex number), then how do I prove that
$${M(jw)}^2 - {N(jw)}^2 = {M(w)}^2 + {N(w)}^2$$
where $j = (-1)^{1/2}$ and $w$ is real?
I am also now really confused as to what exactly is the definition of odd and even functions in the complex domain. 
Kindly help.
As experts say, this seems to be incorrect or atleast incomplete. However, I have taken this from a standard textbook on Network analysis and synthesis by F. F. Kuo. If this helps someone in finding the appropriate conditions under which this is true, kindly help. The book doesn't seem to indicate anything besides those I have mentioned. Thank you for the responses so far.
 A: Here's a counterexample (for the statement as written): $$M(z)=\begin{cases}0 & z\in \Bbb R\\1 & \text{otherwise},\end{cases}$$ $$N(z)=\begin{cases}0 & jz\in \Bbb R\\2 & \text{otherwise}.\end{cases}$$
Then for any $w\in\Bbb R$, we have $$M(jw)^2-N(jw^2)=1-0=1\neq4=0+4=M(w)^2+N(w)^2.$$
See rschwieb's answer for a condition that will be sufficient for your identity to hold.
A: Did you mean to stipulate that they are multiplicative functions?
If $M$ and $N$ are multiplicative, then:
$$M(jw)^2=M((jw)^2)=M(-w^2)=M(w^2)=M(w)^2$$ 
and 
$$N(jw)^2=N((jw)^2)=N(-w^2)=-N(w^2)=-N(w)^2$$.
That would prove your identity.

EDIT
If $M$ and $N$ are polynomials, as you have now stipulated, then $M$ written as a polynomial can only have even powers of the indeterminate, and so $M(jw)^2=M(w)^2$.
EDIT2
A crass mistake led my polynomial counterexample wrong, but now there are already polynomial counterexamples in the other solutions.
A: Polynomial counterexample: $M(x)=x^2+1$, $N(x)=0$.
The equality reduces to
$((ix)^2+1)^2 = (x^2+1)^2$
$(-x^2+1)^2 = (x^2+1)^2$
which is false for $x=1$.

How did I find the counterexample:
Even polynomials have form $M(x)=P(x^2)$ and odd polynomials have form $N(x)=xQ(x^2)$, where $P,Q$ are polynomials.
This gives $P(-x^2)^2+x^2 Q(-x^2)^2 = P(x^2)^2 + x^2 Q(x^2)$.
Setting $Q=0$, we get $P(-x^2)^2 = P(x^2)^2$ where $P$ is any polynomial. We can take any polynomial that does not satisfy this equality, for example $P(x)=x+1$, therefore $M(x)=x^2+1$.
