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Algebra Level 5

If ${\theta}_{1},{\theta}_{2},{\theta}_{3} and {\theta}_{4}$ are four real numbers,then any root of the equation(where $z$ is a complex number):

$sin {\theta}_{1}{z}^{3} + sin {\theta}_{2}{z}^{2} + sin {\theta}_{3}z + sin {\theta}_{4} = 3$

lying inside a unit circle $\left| z \right| = 1$ , always satisfies the inequality :

mod(z) > 3/4 mod(z) > 4/3 mod(z)> 5/6 mod(z) > 2/3

1 solution

Arif Ahmed
Nov 13, 2014

If z satisfies the given equation , then $z \neq 0$ , because that would imply $sin {\theta}_{4} = 3$

Using given equation : |3| = | $sin {\theta}_{1}{z}^{3}+sin {\theta}_{2}{z}^{2}+sin {\theta}_{3}{z}+sin {\theta}_{4}$ |

Therefore , $|sin {\theta}_{1}||{z}^{3}|+|sin {\theta}_{2}||{z}^{2}|+|sin {\theta}_{3}||z|+|sin {\theta}_{4}| \ge 3$

Since $sin x \le 1$ ,

$3 \le |{z}^{3}|+|{z}^{2}|+|z|+1$

$3 < 1+ |z|+|{z}^{2}|+|{z}^{3}|+|{z}^{4}|+. . . . .$

$3 < \frac{1}{1-\left|z\right\|}$

Therefore , $\left|z\right| > \frac{2}{3}$

Done the same way (y) :)

Anandhu Raj - 6 years, 3 months ago