I don't know how to solve this exercise about the roots of a complex polynomial.
Show that the roots of $P(z)=z^9 + iz^5 -400$ has module between $1$ and $2$. How many of them have both real and imaginary part strictly positive?
Hint:study the behavior of $Re(P(z))$ and $Im(P(z))$ in order to find the variations of $arg(P(z))$.
I started writing $z=re^{i\theta}$, $r>0$. The real part $Re(P(z))=r^9cos(9\theta)-r^5sin(5\theta) -400$, while $Im(P(z))=r^9 sin(9\theta)+r^5cos(5\theta)$.
If $z$ is a root, then $\theta=0$, and from $Re(P(r,0))=r^9-400=0$, I got that $1<r<2$.
But I can't understand how to study those functions and how to proceed.
I would brutally write the inequalities $Re(P(z))>0$ and $Im(P(z))>0$, but it doesn't seem to be a good idea.
Any advice?
For $z \in [0;2]$, $\Re(P(z))$ increases from $-400$ to $112$, and $\Im(P(z))$ increases from $0$ to $32$. $P(z)$ passes above the origin, so the argument of $P(z)$ diminishes from $\pi$ to $\arctan(32/112) = a_1$.
Do the same thing for $z \in i[0;2]$, $P(z)$ moves from $-400$ to $-432+512i$, so when you start from $2i$ and go down to $0$, the argument of $P(z)$ increases from $\pi-\arctan(512/432)=a_2$ to $\pi$.
Finally, on the quarter circle from $2$ to $2i$, $P(z)$ is approximately $z^9$, so its argument increases by about $9\pi/2$. The error of the approximation is the combination of the errors at $2$ and $2i$ using straight line paths, so the exact amount is $9\pi/2 + (a_2-\pi/2) - (a_1-0)$
In total on the path $0 \to 2 \to 2i \to 0$, the argument changes by $(a_1-\pi)+(9\pi/2+a_2-\pi/2-a_1)+(\pi-a_2) = 4\pi$, so there are two zeroes in that quadrant.