5
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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?

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

$$z^9(11z+10i)=11-10iz$$

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 $11z+10i$.

$$z^9=\frac{11-10iz}{11z+10i}$$.

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 $|z|=1$.

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  • $\begingroup$ 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
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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 $$\boxed{\text{ on |z|=1}}.$$

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  • $\begingroup$ 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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