Please help. Thx (1 Viewer)

helpfulperson? · Dec 31, 2023

Can someone do part (b)?. Thx

scaryshark09 · Dec 31, 2023

i think this works

liamkk112 · Dec 31, 2023

by conjugate root theorem wbar(w') and 1/wbar (1/w') are also roots
using sum of roots we see w+w' + (w+w')/ww' = -b or 2Re(w) + 2Re(w)/|w|^2 = -b
from here by taking absolute value we can see |b| = |2Re(w)(1+1/|w|^2)|
clearly 1/|w|^2>0 => |b| > |2Re(w)(1+0)| = 2|Re(w)| as required

Luukas.2 · Dec 31, 2023

This question is much more difficult than the above answers suggest, though I suspect that it is due to bad question writing. It highlights a more general issue, however, which is the hazards in making assumptions without noticing them, as they can be unjustified and sometimes render everything that follows as flawed.

Both the above answers have assumed that the four roots must be

$x = \omega,\ \bar{\omega},\ \frac{1}{\omega},\text{ and } \frac{1}{\bar{\omega}}.$

This could be true, but it is not necessarily true.

There is another possibility, which is that $|\omega| = 1.$ In this case:

$|\omega| = 1 \qquad \implies \qquad \frac{1}{\omega} = \frac{\bar{\omega}}{\omega\bar{\omega}} = \frac{\bar{\omega}}{|\omega|^2} = \frac{\bar{\omega}}{1} = \bar{\omega}$

Hence, the two demonstrated roots are themselves a conjugate pair as $\bar{\omega} = \omega^{-1}$ . The remaining two roots could then be:

another conjugate pair; or,
two distinct real roots; or,
a single real double root.

A complete proof requires that these options be investigated and the required result either established as true or the option be demonstrated to be impossible. I suspect that the question means to exclude these possibilities, and so should define $\omega$ as a "non-real root of the equation where $|\omega| \neq 1$ ." Unfortunately, it doesn't. MX2 students need to be very careful with unjustified assumptions and be vigilant in recognising when an assumption is being made.

Looking at one of these options (and leaving the others for others), the double real root case, can be shown to be impossible as the equation could then be factorised as

$\begin{align*} (x - \omega)\left(x - \bar{\omega}\right)(x - k)^2 &= \left[x^2 - \left(\omega + \bar{\omega}\right)x + \omega\bar{\omega}\right]\left(x^2 - 2kx + k^2\right) \qquad \text{for some $k \in \mathbb{R}$} \\ &= \left(x^2 - 2\Re{(\omega)}x + |\omega|^2\right)\left(x^2 - 2kx + k^2\right) \\ & = \left(x^2 - 2\Re{(\omega)}x + 1\right)\left(x^2 - 2kx + k^2\right) \qquad \text{as $|\omega|^2 = 1$} \\ &= x^4 - 2\left(\Re{(\omega)} + k\right)x^3 + \left(1 + 4k\Re{(\omega)} + k^2\right)x^2 - 2k\left(1 + k\Re{(\omega)}\right)x + k^2 \end{align*}$

in which case, by equating constants, $k^2 = 1$ , and by equating coefficients of $x$ and $x^3$ , it follows that

$\begin{align*} b = 2\left(\Re{(\omega)} + k\right) &= -2k\left(1 + k\Re{(\omega)}\right) \\ \Re{(\omega)} + k &= -k - k^2\Re{(\omega)} \\ \left(1 + k^2\right)\Re{(\omega)} &= -2k \\ \Re{(\omega)} &= -\frac{2k}{1 + k^2} \\ &= -\frac{2k}{1 + 1} = -k \qquad \text{since $k^2 = 1$} \\ &= \mp1 \qquad \text{as $k = \pm1$} \\ \\ \text{Hence,} \qquad b &= 2\left(\Re{(\omega)} + k\right) \\ &= 2\left(-k + k\right) \qquad \text{on using $\Re{(\omega)} = -k$} \\ &= 0 \end{align*}$

This cannot be true as $|\omega| = 1$ and $\Re{(\omega)} = \mp1$ requires $\omega = \mp1 \in \mathbb{R}$ and so violates the requirement that $\omega$ be a non-real root of the equation.

In fact, this case gives

$c = 1 + 4k\Re{(\omega)} + k^2 = 1 + 4 \times \pm1 \times \mp1 + 1 = -2$

and so the equation is:

$\begin{align*} x^4 + bx^3 + cx^2 + bx + 1 = 0 \\ x^4 - 2x^2 + 1 &= 0 \qquad \text{as $b = 0$ and $c = -2$} \\ \left(x^2 - 1\right)^2 &= 0 \\ (x + 1)^2(x - 1)^2 &= 0 \end{align*}$

and the equation has only real roots, being two double roots at $x = 1$ and $x = -1$ . The fact that the required identity is only true if $\omega$ is chosen as $\omega = -1$ , yielding

$|b| > 2\Re{(\omega)} \qquad \implies \qquad 0 > -2$

but is untrue if $\omega = 1$ , is chosen as it yields the false statement that

$|b| > 2\Re{(\omega)} \qquad \implies \qquad 0 > 2$

is irrelevant as neither choice of $\omega$ is valid.

I don't know if the claimed result is valid under the other options, or if those options are also impossible, but they need separate investigation to properly established either the truth of the proposition in the question or that the question is invalid without the case of $|\omega| = 1$ being excluded.

Luukas.2 · Dec 31, 2023

Note, also, that @scaryshark09's method can be shortened by keeping $\omega$ and not taking $\omega = p + iq$ :

If $|\omega| \neq 1$ , then summing the roots gives:

$\begin{align*} \text{Let $S$ = Sum of roots} &= \omega + \bar{\omega} + \frac{1}{\omega} + \frac{1}{\bar{\omega}} \\ &= \omega + \bar{\omega} + \frac{\bar{\omega} + \omega}{\omega\bar{\omega}} \qquad \text{on forming a common denominator} \\ &= \left(\omega + \bar{\omega}\right)\left(1 + \frac{1}{|\omega|^2}\right) \qquad \text{on factorising, noting that $\omega\bar{\omega} = |\omega|^2$} \\ &= 2\Re{(\omega)}\left(1 + \frac{1}{|\omega|^2}\right) \qquad \text{as $\omega + \bar{\omega} = 2\Re{(\omega)}$} \\ \\ \text{Hence, the magnitude of $S$} = |S| &> 2\Re{(\omega)} \qquad \text{since $1 + \frac{1}{|\omega|^2} > 1$ for all $\omega \neq 0$} \\ |-b| &> 2\Re{(\omega)} \qquad \text{since the sum of the roots is $\frac{-b}{1}$ from the equation} \\ \\ \text{Thus,} \qquad |b| &> 2\Re{(\omega)} \qquad \text{as required} \end{align*}$

Aeonium · Dec 31, 2023

Luukas.2 said:
This question is much more difficult than the above answers suggest, though I suspect that it is due to bad question writing. It highlights a more general issue, however, which is the hazards in making assumptions without noticing them, as they can be unjustified and sometimes render everything that follows as flawed.

Both the above answers have assumed that the four roots must be

$x = \omega,\ \bar{\omega},\ \frac{1}{\omega},\text{ and } \frac{1}{\bar{\omega}}.$

This could be true, but it is not necessarily true.

There is another possibility, which is that $|\omega| = 1.$ In this case:

$|\omega| = 1 \qquad \implies \qquad \frac{1}{\omega} = \frac{\bar{\omega}}{\omega\bar{\omega}} = \frac{\bar{\omega}}{|\omega|^2} = \frac{\bar{\omega}}{1} = \bar{\omega}$

Hence, the two demonstrated roots are themselves a conjugate pair as $\bar{\omega} = \omega^{-1}$ . The remaining two roots could then be:

another conjugate pair; or,

two distinct real roots; or,

a single real double root.

A complete proof requires that these options be investigated and the required result either established as true or the option be demonstrated to be impossible. I suspect that the question means to exclude these possibilities, and so should define $\omega$ as a "non-real root of the equation where $|\omega| \neq 1$ ." Unfortunately, it doesn't. MX2 students need to be very careful with unjustified assumptions and be vigilant in recognising when an assumption is being made.

Looking at one of these options (and leaving the others for others), the double real root case, can be shown to be impossible as the equation could then be factorised as

$\begin{align*} (x - \omega)\left(x - \bar{\omega}\right)(x - k)^2 &= \left[x^2 - \left(\omega + \bar{\omega}\right)x + \omega\bar{\omega}\right]\left(x^2 - 2kx + k^2\right) \qquad \text{for some $k \in \mathbb{R}$} \\ &= \left(x^2 - 2\Re{(\omega)}x + |\omega|^2\right)\left(x^2 - 2kx + k^2\right) \\ & = \left(x^2 - 2\Re{(\omega)}x + 1\right)\left(x^2 - 2kx + k^2\right) \qquad \text{as $|\omega|^2 = 1$} \\ &= x^4 - 2\left(\Re{(\omega)} + k\right)x^3 + \left(1 + 4k\Re{(\omega)} + k^2\right)x^2 - 2k\left(1 + k\Re{(\omega)}\right)x + k^2 \end{align*}$

in which case, by equating constants, $k^2 = 1$ , and by equating coefficients of $x$ and $x^3$ , it follows that

$\begin{align*} b = 2\left(\Re{(\omega)} + k\right) &= -2k\left(1 + k\Re{(\omega)}\right) \\ \Re{(\omega)} + k &= -k - k^2\Re{(\omega)} \\ \left(1 + k^2\right)\Re{(\omega)} &= -2k \\ \Re{(\omega)} &= -\frac{2k}{1 + k^2} \\ &= -\frac{2k}{1 + 1} = -k \qquad \text{since $k^2 = 1$} \\ &= \mp1 \qquad \text{as $k = \pm1$} \\ \\ \text{Hence,} \qquad b &= 2\left(\Re{(\omega)} + k\right) \\ &= 2\left(-k + k\right) \qquad \text{on using $\Re{(\omega)} = -k$} \\ &= 0 \end{align*}$

This cannot be true as $|\omega| = 1$ and $\Re{(\omega)} = \mp1$ requires $\omega = \mp1 \in \mathbb{R}$ and so violates the requirement that $\omega$ be a non-real root of the equation.

In fact, this case gives

$c = 1 + 4k\Re{(\omega)} + k^2 = 1 + 4 \times \pm1 \times \mp1 + 1 = -2$
and so the equation is:

$\begin{align*} x^4 + bx^3 + cx^2 + bx + 1 = 0 \\ x^4 - 2x^2 + 1 &= 0 \qquad \text{as $b = 0$ and $c = -2$} \\ \left(x^2 - 1\right)^2 &= 0 \\ (x + 1)^2(x - 1)^2 &= 0 \end{align*}$

and the equation has only real roots, being two double roots at $x = 1$ and $x = -1$ . The fact that the required identity is only true if $\omega$ is chosen as $\omega = -1$ , yielding

$|b| > 2\Re{(\omega)} \qquad \implies \qquad 0 > -2$

but is untrue if $\omega = 1$ , is chosen as it yields the false statement that

$|b| > 2\Re{(\omega)} \qquad \implies \qquad 0 > 2$

is irrelevant as neither choice of $\omega$ is valid.

I don't know if the claimed result is valid under the other options, or if those options are also impossible, but they need separate investigation to properly established either the truth of the proposition in the question or that the question is invalid without the case of $|\omega| = 1$ being excluded.

bad qn is decent possibility considering kurt rips a lot of qns from elsewhere (as most tutors/teachers do)

Luukas.2 · Dec 31, 2023

Aeonium said:
bad qn is decent possibility considering kurt rips a lot of qns from elsewhere (as most tutors/teachers do)

Bad questions can provide good teaching points... and good educators fix questions when they find problems.

Please help. Thx (1 Viewer)

helpfulperson?

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scaryshark09

∞∆ who let 'em cook dis long ∆∞

liamkk112

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Luukas.2

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Luukas.2

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Aeonium

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Luukas.2

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