Prove the roots of a complex polynomial are imaginary For an equation $f(z) = z^5 - 6z^4 + 15z^3 - 34z^2 +36z -48$ show that roots $f(z) = 0$ of this equation include 2 purely imaginary roots, and find them. 
I thought to substitute in $z=x+iy$ to show that you can only get a solution for $y$s, but that seems like a really long and complicated process. I then thought to use $z=re^{i\theta}$ but I'm not sure how I'd use this to get the roots from the resultant equation.
 A: Note that if you have a pair of conjugate imaginary roots you have a factor of the form $(z+ai)(z-ai)=z^2+a^2$. Here $a$ need not be an integer, of course, so we look for a factor of the form $z^2+b$ with $b$ a positive real number by substituting $z^2=-b$ in the original equation to obtain $$b^2z-6b^2-15bz+34b+36z-48=0$$
We then note that this splits as $$z(b^2-15b+36)-(6b^2-34b+48)=0$$
If we are looking for real $b$ we know that $z$ is imaginary for our roots so both expressions in $b$ must vanish. We find a common factor $b-3$ - the common factor here shows that there is a pair of purely imaginary roots. This gives us a factor $z^2+3$.
A: This fifth degree polynomial has 5 roots (counting multiplicities), one of them real an the others complex conjugates. To rule out multiple roots, compute the greatest common divisor with its derivative, which is 1. So it has 5 different roots.
By Descartes' rule of signs it has at most 5 positive roots (i.e, it could have 1, 3, or 5 positive roots), and at most 0 negative ones.
The rational root test tells you possible rational roots are divisors of 48. Sure enough, $p(4) = 0$.
My tame computer algebra system (maxima) tells me $p(z) = (z - 4) (z^2 + 3) (z^2 - 2 z + 4)$.
Yes, it's got two pure imaginary roots.
You could have found this out by trial division by $z^2 + a$, adjusting $a$ so the division goes through...
A: Your polynomial $f$ has real coefficients. Therefore, if $r$ is one root of $f$, $\overline r$ will be another. If $r$ is also imaginary, then $\overline r = -r$. Thus if there is an imaginary root $r$ of $f$, then we must have $f(r)=f(-r)=0$, in other words, the polynomials $f(x)$ and $f(-x)$ have at least one common root, namely $r$.
Therefore, calculate the GCD of $f(x)$ and $f(-x)$, since it must contain all of these common roots. This turns out to be $x^2+3$, whose roots are $i\sqrt 3$ and $-i\sqrt 3$. Therefore these are roots of $f$.
A: Let's consider the function
$$g(z) = f(iz) = iz^5 - 6z^4 -15iz^3 + 34 z^2 + 36iz - 48$$
Then, $f$ has a pure imaginary root if (and only if) $g$ has a real root.  But, if $g$ has a real root, then the polynomials
$$R(z) = \operatorname{Re}(g(z)) = -6z^4 + 34 z^2 - 48$$
$$I(z) = \operatorname{Im}(g(z)) = z^5 - 15 z^3 + 36z = z(z^4 - 15 z^2 + 36)$$
must have this same real root.  Thus, we look for common roots of $R$ and $I$.
Since $R(z)$ is quadratic in $z^2$, any root $z$ satisfies
$$z^2 = \frac{-34 \pm \sqrt{34^2 - 4(-6)(-48)}}{2(-6)} = \frac{34 \pm 2}{12} = 3, \frac{5}{2}.$$
so, the roots of $I(z)$ are
$$z = \pm \sqrt{3}, \pm \sqrt{\frac{5}{2}}.$$
Putting these values in for $z$ in $I(z)$, we find that $z= \pm\sqrt{3}$ are also roots of $I(z)$ (and $\pm \sqrt{\frac{5}{2}}$ are not roots), so
$$f(\pm i\sqrt{3}) = g(\pm\sqrt{3}) = R(z) + iI(z) = 0$$
and $f$ has two purely imaginary roots.
