# 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.

-
The definition of odd and even functions is exactly the same. $f:\Bbb C\to\Bbb C$ is even if $f(-z)=f(z)$ for all $z$, and odd if $f(-z)=-f(z)$ for all $z$. – Cameron Buie Nov 15 '12 at 15:24
Do you have any assumptions on $M,N$ other than even/oddness? LHS depends only on values of $M,N$ at imaginary values and RHS on values of $M,N$ at real values. Doesn't seem to be true. – sdcvvc Nov 15 '12 at 15:26
$M$,$N$ both are polynomials in $s$. Does that help? – Guanidene Nov 15 '12 at 15:29
I agree with sdcvvc: it sounds like you are leaving something out. I speculated that it was multiplicativity in my answer, since that more or less obviously leads to the solution. – rschwieb Nov 15 '12 at 15:31
@Guanidene Well yes that would certainly help... – rschwieb Nov 15 '12 at 15:31

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.

-

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

$$N(jw)^2=N((jw)^2)=N(-w^2)=-N(w^2)=-N(w)^2$$.

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.

-

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 $M(x)=x^2+1$.

-