If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is

If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is

$\bf{My\; Try::}$ Given $11z^{10}+10iz^9+10iz-11 = 0\Rightarrow \displaystyle z^9 = \frac{11-10iz}{11z+10i}.$

Now Put $z = x+iy\;,$ we get $\displaystyle (x+iy)^9 = \frac{11-10i(x+iy)}{11(x+iy)+10i} = \frac{(11+10y)-10ix}{11x+i(10+11y)}$

Now i did not understand how can i solve it,

Help Required

Thanks

Apply the consequence of the argument principle known as Rouché's theorem. Hope you have learnt it.

Take the function $g(z) = 11 z^{10} + 10iz^9 + 10 iz -11$. It is a polynomial of degree 10 and so it is analytic and the equation $g(z) = 0$ has 10 roots.

Consider $f(z) = 11 z^{10} -11$.

Now see $|g(z) - f(z)| = |10(z^9 - z)| < f(z)$ on the circle $|z| = r$. You may take $r = 1.1$. So $g$ and $f$ will have same number of zeros inside $|z| <r$.

See $f(z) = 0$ has 10 roots on the circle $|z| = 1$. Do it now.

After the substitution $x=iy$ , we get $11y^{10}+10y^9+10y+11=0$. Now, as $y=0$ is not a root of this equation, dividing both sides by $y^5$ gives $11y^5+11/y^5+10y^4+10/y^4=0$. Now put$t=y+1/y$ to get $11t^5+10t^4-55t^3-40t^2+55t+20=0$. You can plot the expression in $t$ or use calculus to see that the equation has 5 real roots. So, every value of $t$ which satisfies the equation is real. As $t=y+1/y$, $y+1/y$ is also real. Taking $y=re^{i \theta}$ you can see that $r$ must be 1 for$y+1/y$ to be real. As $r$ is the magnitude of $y$, the roots have unit modulus.