Question about complex analysis; writing $f(iy) = u(y) + iv(y)$ 
Let $f(z) = z^n + a_{n-1}z^{n-1} + ... $ be a monic polynomial of
  degree $n$, and assume that $f(iy) \ne 0$ for all $y \in \mathbb R$.
  Write $f(iy) = u(y) + iv(y)$ and express the number of roots of $f$ in
  the right half plane {$\Re(z)> 0$} in terms of the number and mutual
  position of the real roots of $u$ and $v$.

Now we have the first constant is equal to 1 and I searched and got the idea that it is always possible to write $f(z)=u(x,y)+iv(x,y)$ in here. But I do not know where to start in this question. Any help is appreciated!
 A: You may look at the argument of $f(z)$ as $z$ goes along the boundary of a large half disk in the right half plane.
Thus you could consider the two contours:
$$z=R e^{i\theta}, -\pi/2\leq \theta\leq \pi/2$$
and
    $$ z=-it , \ -R\leq t \leq R .$$
For large enough $R$ the argument of $f(z)$ along the first contour will increase with $\pi n$.
The intertwined ordering of the real zeros (counting multiplicity) of $u$ and $v$ will tell you how the argument of $f(z)$ changes as you go along the imaginary axis. For example a zero of $v$ in between two zeros of $u$ tell you that the argument has increased (or decreased) by $\pi$. But for two consecutive zeros of $u$ there is no argument change but a reversal of the orientation. 
It seems to me, however, that there is an information missing. For example, the polynomials $z^2\pm z+1$ will have the same locations for the real zeros 
of $(u,v) = (-y^2+1,\pm y)$ but two roots in the right half plane in one case and 0 in the other. The point is that you need to know the argument change at some point e.g. by knowing the signs of both $u$ and $v$ at, say $y=R$, but the given information only allows you to calculate one of them. Consequently an argument computation based on the location of real roots of $u$ and $v$ will give you either the number of roots in the right or the left half plane, but you can not tell which.
