Let $\omega$ be the complex number satisfying $\omega^3 = 1$ and $\text{Im}(\omega) > 0$ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2008 - Paper 1
Question 6
Let $\omega$ be the complex number satisfying $\omega^3 = 1$ and $\text{Im}(\omega) > 0$. The cubic polynomial, $p(z) = z^3 + az^2 + bz + c$, has zeros $1, -\omega$ ... show full transcript
Worked Solution & Example Answer:Let $\omega$ be the complex number satisfying $\omega^3 = 1$ and $\text{Im}(\omega) > 0$ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2008 - Paper 1
Step 1
Find $p(z)$
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Answer
To find the polynomial p(z), we first recognize that the zeros of the polynomial are 1, −ω, and −ω. Since ω satisfies ω3=1, we can express these zeros in terms of ω:
Calculate ω=ω1, since for a complex number on the unit circle, the conjugate is the reciprocal.
Therefore, the zeros can be written as 1, −ω, and −ω1.
The polynomial can thus be formed as:
p(z)=(z−1)(z+ω)(z+ω1)
Expand this product:
p(z)=(z−1)(z2+z(ω+ω1)+1)
Finding value of ω+ω1=ωω2+1=2cos(θ) for some angle θ where \text{Im}(ω)>0
Thus, we have:
p(z)=(z−1)(z2+2cos(θ)z+1)
The result can be simplified further and for specific values of a, b, and c, will yield the final form of the cubic polynomial.
Step 2
Show that the line $\ell$ has equation $bx \sec \theta - ay \tan \theta - ab = 0$
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To show that the equation of the line ℓ is given by:
Start from the known properties of the hyperbola and the point P(secθ,btanθ).
The derivative at point P gives the slope of the tangent line ℓ:
The slope of a tangent to a hyperbola is given by implicit differentiation.
Substitute the coordinates of point P into the equation of the line:
y−btanθ=m(x−secθ)
Where m is the slope derived above.
4. Rearranging gives the desired result:
bxsecθ−aytanθ−ab=0.
Step 3
Show that $SR = \frac{ab(\sec\theta - 1)}{\sqrt{a^2 \tan^2\theta + b^2 \sec^2 \theta}}$
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We know from the geometry of the hyperbola that SR is the distance from point S to the line ℓ.
The distance from a point to a line can be expressed as:
d=A2+B2∣Ax+By+C∣
Using coefficients from the line equation.
3. Substituting S(−ae,0) and calculating gives the necessary distance formula for SR.
Step 4
Show that $SR \times S'R' = b^2$
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From the previous results, express S′R′ in a similar manner as SR using S′.
Since SR and SR′ are equal by definition of hyperbola properties, we can substitute accordingly.
The geometric property can be simplified and shown to equal b2.
Therefore, we conclude:
SR×S′R′=b2.
Step 5
Show that $\frac{1}{r} = \frac{r - 1}{r} [\frac{1}{n} - \frac{1}{r - 1}]$
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Begin with the left hand side:
r1
Rewrite the right hand side:
=rr−1(n1−r−11)
By simplifying, establish the equality through common denominators and properties of fractions.
Step 6
Hence show that, if $m$ is an integer with $m \geq r$, then \[ \frac{1}{r} + \cdots + \frac{1}{r} = \frac{m}{r} - \frac{1}{n} [\frac{1}{r} - \frac{1}{r - 1}] \text{ (total } m \text{ terms)}. \]
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Consider the sum of m terms of r1.
This gives:
rm
Now apply the initial condition from part (i) concerning n and rewrite:
n1(m−r−1m)
Ensure terms are combined properly to yield the final expression.
Step 7
What is the limiting value of the sum \[ \sum_{r=1}^{m} \frac{1}{n} \] as $m$ increases without bound?
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The limiting value evaluates the series:
∑r=1mn1=nm
As m approaches infinity, this implies:
limm→∞nm=∞
Thus, the limiting value of the sum as m increases indefinitely is:
