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.
Best Answer
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 $f$.