<s>[INST] 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
 [/INST] 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.
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