$i^2=-1$

We can compare the real and imaginary part of the equation separately:

Real: $\boxed{11z^{10}-11=0}$ Imaginary: $\boxed{10z^9+10z=0}$

From the real part we get: $11z^{10}-11=0 \Leftrightarrow z^{10}=1 \Leftrightarrow |z|^{10}e^{10i\phi}=1$

Since $e^{10i\phi}$ has always an absolute value of $1$ we can directly conclude that $|z|^{10}=1\Rightarrow\boxed{|z|=1}$

At this point the problem is solved. But we can, to check our answer, do the same thing for the imaginary part as well:

$10z^9+10z=0\Leftrightarrow z(10z^8+10)=0$ Now we have two options: Ether $z=0$ (which makes no sense) or $10z^8+10=0\Leftrightarrow \boxed{z^8=-1}$

We can see clearly that the solution must be a point on the unit-circle and therefore $\boxed{|z|=1}$

(Solution fully completed 2/2/2018.) Let $z$ be a root, i.e., $11z^{10}+10iz^9+10iz-11=0$ . Then take the conjugate of the equation and divide by $\bar{z}^{10}$ (noting $\bar{z}\neq 0$ ). This leads to $11-10i\frac{1}{\bar{z}}-10i\frac{1}{\bar{z}^9}-11\frac{1}{\bar{z}^{10}}=0$ . Adding the equations results in $11\Big(z^{10}-\frac{1}{\bar{z}^{10}}\Big)+10i\Big(z^9-\frac{1}{\bar{z}^9}\Big)+10i\Big(z-\frac{1}{\bar{z}}\Big)=0$ . Multiply by $\bar{z}^{10}$ to obtain $11(|z|^{20}-1)+10i\bar{z}(|z|^{18}-1)+10\bar{z}^9(|z|^2-1)=0$ . Then factorize as $(|z|^2-1)(11(|z|^{18}+|z|^{16}+...+1)+10i\bar{z}(|z|^{16}+|z|^{14}+...+1)+10i\bar{z}^9)=0$ . Then by the triangle inequality, $|11(|z|^{18}+|z|^{16}+...+1)+10i\bar{z}(|z|^{16}+|z|^{14}+...+1)+10i\bar{z}^9|$ $\geq 11(|z|^{18}+|z|^{16}+...+1)-10|z|(|z|^{16}+|z|^{14}+...+1)-10|z|^9$ . Now, I will prove $11(|z|^{18}+|z|^{16}+...+1)-10|z|(|z|^{16}+|z|^{14}+...+1)-10|z|^9>0$ . Assume $|z|=0$ . Then the expression is equal to $11$ , which is greater than $0$ . Next, assume $0<|z|\leq 1$ . Then I can break up, reorder, and bracket the terms as $(11|z|^{18})+(11|z|^{16}-10|z|^{17})+(11|z|^{14}-10|z|^{15})+...(1-|z|)+(10-10|z|^9)$ . It is easily seen that the whole expression is strictly positive. Now assume $|z|>1$ . Write the expression as $(10|z|^{18}-10|z|^{17})+(10|z|^{16}-10|z|^{15})+...+(|z|^{18}+|z|^{16}+...+11-10|z|^9)$ . Each bracketed term is positive. (To see that $|z|^{18}+|z|^{16}+...+11>10|z|^9$ for all $|z|>1$ , first note that the inequality is true for $|z|=1$ . Next, differentiate both sides of the inequality (w.r.t. $|z|$ ) to get a new inequality, which if is true for $|z|>1$ , then the original inequality is true. This leads you to differentiate the inequality again and come to the same conclusion. Repeat the process until you have differentiated the inequality 9 times. This results in $18\cdot 17\cdot ... \cdot 10|z|^9+...+10!|z|>10!$ , which is clearly true for $|z|>1$ . Noting that, in each case, the inequalities are satisfied for $|z|=1$ , this dominoes up to $|z|^{18}+|z|^{16}+...+1>10|z|^9$ to show that it is true for all $|z|>1$ .) Finally, since $|11(|z|^{18}+|z|^{16}+...+1)+10i\bar{z}(|z|^{16}+|z|^{14}+...+1)+10i\bar{z}^9|>0$ , we have $11(|z|^{18}+|z|^{16}+...+1)+10i\bar{z}(|z|^{16}+|z|^{14}+...+1)+10i\bar{z}^9\neq 0$ , so we can divide by it in $(|z|^2-1)(11(|z|^{18}+|z|^{16}+...+1)+10i\bar{z}(|z|^{16}+|z|^{14}+...+1)+10i\bar{z}^9)=0$ to conclude $|z|^2-1=0$ or, equivalently, $|z|=1$ .