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
 A: 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.
A: 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.
A: Let $f(z)=11z^{10}+10iz^9$, $g(z)=(10iz-11)$ and $p=f+g$. For $\theta\in \Bbb R$ we have $$\big|f(e^{i\theta})\big|=|11\cos\theta+11i\sin \theta+10i|$$$$=\sqrt{221+220\sin\theta}$$$$=|10i\cos\theta-10\sin\theta-11|=\big|g(e^{i\theta})\big|.$$
Hence, $\big|p\big|_{\Bbb S^1}-g\big|_{\Bbb S^1}\big|=\big|f\big|_{\Bbb S^1}\big|=\big|g\big|_{\Bbb S^1}\big|$.
Now, for all $\varepsilon>1$ consider $p_\varepsilon(z)=11z^{10}+10iz^9+\varepsilon(10iz-11)$ and $g_\varepsilon(z)=\varepsilon(10iz-11)$. Then, $\big|p_\varepsilon(z)-g_\varepsilon(z)\big|=\big|f(z)\big|<\varepsilon \big|g(z)\big|=\big|g_\varepsilon(z)\big|$ for all $z\in \Bbb S^1$. By Rouche's theorem $p_\varepsilon$ and $g_\varepsilon$ have same number of zeros inside $\Bbb S^1$. But, the only zero of $g_\varepsilon$ lies outside of $\Bbb D$. Hence, every zero of $p_\varepsilon$ lies in $\{z\in\Bbb C:|z|\geq 1\}$. Notice the coefficients of $p_\varepsilon$ converge to coefficients of $p$ as $\varepsilon\to 1+$. Hence, all roots of $p$ also lie in $\{z\in\Bbb C:|z|\geq 1\}$. See Proposition 5.2.1(b) of Artin's algebra.
Similarly, every zero of $q(w):=11+10iw+10iw^9-11w^{10}$ lies in $\{w\in\Bbb C:|w|\geq 1\}$. But, $q(w)=0\text{ for some }w\text{ with }|w|\geq 1\implies p\big(\frac{1}{w}\big)=0$ and conversely, $p(z)=0\text{ for some }z\text{ with }|z|\geq 1\implies q\big(\frac{1}{z}\big)=0$. Now, $p^{-1}(0)\subseteq\{z\in\Bbb C:|z|\geq 1\}$. Hence, we are done.
