Complex number - prove an inequality

Asked Jan 14 '15 at 16:31

Active Jan 14 '15 at 16:53

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Question:

Given that:$$z^n\tan\theta_0 + z^{n-1}\tan\theta_1 + z^{n-2}\tan\theta_2 + ... + \tan\theta_n = 3$$ And that $\theta_i \in (0, \frac{\pi}{4})$, prove that: $$|z| > \frac{2}{3}$$

Approach: I tried to solve the question using $|z_1 + ... + z_n| \leq |z_1| + ... + |z_n|$, however, two issues came up. First of all, the presence of the equality, and second of all the presence of the tan function. If we assume that $\theta$ is ${\pi}\over4$ for all of them, then I am able to get a G.P. Using the formula for the sum of a GP, I get: $$|z| \geq \frac{2}{3}$$ Don't understand how to eliminate the inequality, and unsure if I can take the assumption for the tan function.

edited Jan 14 '15 at 16:53

janmarqz

10,538

asked Jan 14 '15 at 16:31

Gummy bears

3,408

Hint: Triangle inequality and $\sum\limits_{k=0}^{\infty} (2/3)^k = 3$ :) – r9m Jan 14 '15 at 16:38
@r9m Sorry, didn't understand anything. I already used the triangle inequality but don't get the second hint. – Gummy bears Jan 14 '15 at 16:39
It's a geometric sum. – DaveNine Jan 14 '15 at 16:43
@Gummybears consider to the contrary $|z| \le \frac{2}{3}$, then $3 = |\sum\limits_{k=0}^{n} \tan \theta_k z^k| \le \sum\limits_{k=0}^{n} |\tan \theta_k| |z|^k < \sum\limits_{k=0}^{n} (2/3)^k < \sum\limits_{k=0}^{\infty} (2/3)^k = 3$ a contradiction. (since, $|\tan \theta_k| < 1$) – r9m Jan 14 '15 at 16:43
@r9m Please post as an answer with more explanation. I still don't understand. – Gummy bears Jan 14 '15 at 16:45
1

@r9m It's $<$, not $\le$, when passing from the finite sum to the series. – egreg Jan 14 '15 at 16:45
@egreg But why is that? – Gummy bears Jan 14 '15 at 16:46
@egreg right!thanks:) – r9m Jan 14 '15 at 16:47

Complex number - prove an inequality

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