If $θ_1,θ_2,θ_3,θ_4$ are four real numbers, then any root of the equation


lying inside the unit circle $\vert z\vert$=1, satisfies which inequality?

A)$\vert z\vert$ < $\frac{2}{3}$

B)$\vert z\vert$ > $\frac{2}{3}$

C)$\vert z\vert$ < $\frac{1}{2}$

D)None of these

We've recently passed equations of degree 3,4 and I know that it's maximum 4 roots that satisfy the equation. What does it mean for them to lie in the unit circle $\vert z\vert$=1? I guess it has to do something with complexical numbers?

  • 2
    $\begingroup$ Thank you for the edit, Jackie! $\endgroup$
    – amWhy
    Nov 23 '14 at 18:00
  • 1
    $\begingroup$ @amWHy lol you're welcome $\endgroup$ Nov 23 '14 at 18:26
  • $\begingroup$ All your questions are "beautiful"? $\endgroup$
    – JohnD
    Nov 23 '14 at 20:42
  • 2
    $\begingroup$ @John D: it's mathematics; beautiful by definition! Cheers! $\endgroup$ Nov 23 '14 at 20:47
  • 2
    $\begingroup$ Complexical numbers are especially beautistic. $\endgroup$
    – TonyK
    Nov 23 '14 at 21:03

I am assuming that by $\sin \theta_1 z^3$ our OP Jackie means $z^3 \sin \theta_1$ and so forth; with this understanding, we have the given equation

$z^3 \sin \theta_1 + z^2 \sin \theta_2 + z \sin \theta_3 + \sin \theta_4 = 3; \tag{1}$

taking absolute values and using the triangle inequality (several times) yields

$3 = \vert 3 \vert \le \vert z^3 \sin \theta_1 \vert + \vert z^2 \sin \theta_2 \vert + \vert z \sin \theta_3 \vert + \vert \sin \theta_4 \vert; \tag{2}$

now if we repeatedly apply the equality $\vert xy \vert = \vert x \vert \vert y \vert$, $x, y \in \Bbb C$ to (2) we obtain

$3 \le \vert z \vert^3 \vert \sin \theta_1 \vert + \vert z \vert^2 \vert \sin \theta_2 \vert + \vert z \vert \vert \sin \theta_3 \vert + \vert \sin \theta_4 \vert. \tag{3}$

We have $\vert \sin \theta_i \vert \le 1$, $ 1 \le i \le 4$; thus if $\vert z \vert \le 2/3$, (3) yields

$3 \le (\dfrac{2}{3})^3 + (\dfrac{2}{3})^2 + \dfrac{2}{3} + 1; \tag{4}$

however, the sum of the first three terms on the right of (4) is

$\dfrac{8}{27} + \dfrac{4}{9} + \dfrac{2}{3} =\dfrac{38}{27} < 2, \tag{5}$

so (4) becomes

$3 \le \dfrac{38}{27} + 1< 2 + 1 = 3. \tag{6}$

(6) is clearly not happening; this contradiction shows we must have

$\vert z \vert > \dfrac{2}{3}; \tag{7}$

thus (B) is the correct choice.

It is worth observing that with $\theta_i = \pi/2$, $1 \le i \le 4$, there is in fact a real solution $r$ to (1) satisfying $2/3 < r < 1$; this may be seen by applying the intermediate value theorem to the polynomial

$p(r) = \sum_0^3 r^i, \tag{8}$

which satisfies $p(2/3) = 65/27 < 3$, $p(1) = 4$. This shows that for at least some values of $\theta_i$, there is indeed a solution $z$ to (1) with $\vert z \vert < 1$; the problem is not vacuous.

As a closing comment, and to address Jackie's last two questions, we recall that for $z = a + bi$, $\vert z \vert = \sqrt{a^2 + b^2}$; saying $z$ lies within the unit circle thus means either $a^2 + b^2 < 1$ or $a^2 + b^2 \le 1$, depending on exaxtly how one parses the word "within"; this indeed has a lot to do with "complexical numbers".

Hope this helps. Cheers,

and as ever

Fiat Lux!!!


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