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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

Greg Martin
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juantheron
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  • gain intuition at [ http://www.wolframalpha.com/input/?i=solve++11z%5E10%2B10+i+z%5E9%2B10+i+z%E2%88%9211%3D%3D0%2C+z ] – janmarqz Feb 09 '14 at 04:30
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    Perhaps rewriting as $-11(iz)^{10} + 10(iz)^9 + 10(iz) - 11 = 0$ or $11(iz)^{10} - 10(iz)^9 - 10(iz) + 11 = 0$ might help. – Yiyuan Lee Feb 09 '14 at 04:33
  • It asks for the possible values of $|z|$ for a reason -- all of the complex roots lie on the same circle (i.e. have the same magnitude) – MT_ Feb 09 '14 at 04:44
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    Show that $w=i(z-z^{-1})$ satisfies a quintic. If you can show this quintic has five real roots for $w$, then show that $z$ lies on a circle. – Empy2 Feb 09 '14 at 06:25

3 Answers3

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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.

Did
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Supriyo
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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.

Caleb Stanford
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Ilovemath
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    Correction: $r$ must be $1$ or $y$ must be real (if $y$ is real, $y + \frac{1}{y}$ is real). The latter can be ruled out by determining that $11y^{10}+10y^{9}+10y+11$ has no real roots. – Caleb Stanford Dec 04 '21 at 23:07
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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.

Sumanta
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