Spring 2014 - Problem 12

Solution.

Notice that if $a$ is a root of $f\p{z}$, then so is $-a$. Moreover, if $\Im{z} = 0$, then write $z = t \in \R$ which gives

which has strictly positive real part, so $f$ does not vanish on the real line. Hence, $a \neq -a$ and the zeroes of $f$ come in pairs $\p{a, -a}$, and if $\Im{a} > 0$, then $\Im{a} < 0$. By the fundamental theorem of calculus, $f$ has precisely $6$ roots, and half of them must lie in the upper half-plane, so $f$ has $3$ roots in the upper half-plane.