If $z$ is a root of the equation $$11z^{10}+10iz^9+10iz-11=0$$ Find value of $|z|$
I assumed the root as $z=re^{it}$ We get:
$$11r^{10}\cos(10t)-10r^9\sin(9t)-10r\sin (t)-11=0 \tag{1}$$ and $$11r^9\sin(10t)+10r^8\cos(9t)+10\cos(t)=0 \tag{2}$$
Any way from here?
One way to do it is to show by Rouche that all critical points are in the closed unit disc - this is immediate since $P'(z)=110z^9+90iz^8+10i$ and $110>90+10$ so $|P'(z)-110z^9|<|110z^9|$ on the unit circle etc
But now $P$ is self inversive since if $z$ is a root, $\frac{1}{\bar z}$ is a root and then it is an easy exercise to show that a self inversive $P$ must have all the roots on the unit circle when all its critical points are inside the closed unit disc so $|z|=1$
($\Re{\frac{zf'}{f}}>0, 1<|z|<1+\epsilon$ and the argument principle)
I´m not sure, but if you try to resolve $11\left(z^{10}-1\right)+10i\left(z^9+1\right)=0$ maybe it´s gonna be more easy.
