The question is as follows:

All the roots of the equation $11z^{10}+10iz^9+10iz-11=0$ lie:
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (i=\sqrt{-1})$
(a) inside $|z|=1$
(b) on $|z|=1$
(c) outside $|z|=1$
(d) can't say

The answer is (b). I tried factorizing it, but to no avail. Also, it doesn't appear to me that taking modulus would help. How to approach this problem?


Keeping the higher order terms on the LHS and the lower order terms on the RHS, we have


We first check that $11z+10i \neq 0$. Suppose on the contrary that it is, then $z=\frac{-10i}{11}$ and $LHS=0$, but $RHS=11-10i\left(\frac{-10i}{11}\right) \neq 0$. Hence, we can divide both sizes by $11z+10i$.


Taking modulus and squaring both sides,

\begin{align} \left|z\right|^{18}&=\frac{\left|11-10iz\right|^2}{\left|11z+10i\right|^2}\\ &=\frac{\left(11-10iz \right)\left(11+10i\bar{z} \right)}{\left(11z+10i \right)\left( 11\bar{z}-10i \right)}\\ &=\frac{121+110i\bar{z}-110iz+100\left| z\right|^2}{121 \left|z \right|^2+110i\bar{z}-100iz+100}\\ &= \frac{A}{B} \end{align}

where I let the numerator expression in the second last line above be $A=121+110i\bar{z}-110iz+100\left| z\right|^2$ and the denominator in the second last line above be $B=121 \left|z \right|^2+110i\bar{z}-100iz+100.$

If $|z|>1$, then $|z|^{18}>1$, then $A>B$ and $A-B>0$, but $$A-B=121 \left( 1-\left| z\right|^2\right)+100\left( \left| z\right|^2-1\right)=21(1-\left| z\right|^2)$$

which is negative since $\left|z\right|^2>1$ by our assumption. Hence a contradiction.

Similarly, if $|z|<1$, $|z|^{18}<1$, then $A<B$ and $A-B<0$, but $A-B=21(1-|z|^2)>0$ which is again a contradiction.

Hence $|z|=1$.

  • $\begingroup$ That's a clever solution! Thank you! $\endgroup$ – FreezingFire Aug 31 '16 at 18:43
  • $\begingroup$ @SiongThyeGoh Shouldn't it be: $\frac{\left(11-10iz \right)\left(11+10i\bar{z} \right)}{\left(11z+10i \right)\left( 11\bar{z}-10i \right)}\\$ $=\frac{121+110i\bar{z}-110iz+100\left| z\right|^2}{121 \left|z \right|^2+110i\bar{z}-110iz+100}\\$? $\endgroup$ – Tapi Aug 4 at 10:37
  • $\begingroup$ hi, somehow I can't tell the difference. $\endgroup$ – Siong Thye Goh Aug 4 at 10:49

Substitution $z=e^{it}$ gives the trigonometrical equation $$11\sin5t+10\cos4t=0,\qquad(1)$$ or $$\cos 4t=-1.1\sin 5t.$$ Easy to see that $$RHS\left(\dfrac{2k+1}{10}\pi\right)=1.1(-1)^{k+1}$$ for $k=-3,-2,-1,0,1,2$, so LHS and RHS have at least five intersections for $t\in\left(-\dfrac\pi2,\dfrac\pi2\right)$.

enter image description here

This means that $(1)$ has at least 5 real roots for $t\in\left(-\dfrac\pi2,\dfrac\pi2\right)$.
On the other hand, it is known that $$\sin5t=16\sin^5t-20\sin^3t+5\sin t$$ and $$\cos4t=8\sin^4t-8\sin^2t+1,$$ so $(1)$ is equivalent to $$176y^5+80y^4-220y^3-80y^2+55y+10=0,\quad y=\sin t.$$ In this way, the 5th order polynomial has $5$ real roots for $y\in(-1,1)$.
Therefore, equation $(1)$ has only real roots.

Thus, the right answer is $$\boxed{\text{ on |z|=1}}.$$

  • $\begingroup$ How do you show that your trig equation has only real roots? $\endgroup$ – DanielWainfleet Aug 29 '16 at 20:33
  • $\begingroup$ @user254665 Proved... $\endgroup$ – Yuri Negometyanov Aug 29 '16 at 21:35
  • $\begingroup$ @Dr.MV Proved.. $\endgroup$ – Yuri Negometyanov Aug 29 '16 at 21:36
  • $\begingroup$ @Dr.MV. The substitution $ z=e^{it}$ does not assume that $t$ is a real number. So it does not assume $|z|=1.$ $\endgroup$ – DanielWainfleet Aug 29 '16 at 21:40
  • $\begingroup$ @Dr.MV For the complex $t$ used exponential substitution does not reduce the community. But we proved that $t$ is real. What's wrong? $\endgroup$ – Yuri Negometyanov Aug 29 '16 at 21:41

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