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 $1, -\omega$, and $-\overline{\omega}$. Find $p(z)$. Let $P(\sec \theta, b \tan \theta)$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ where $a > 0$ and $b > 0$ as shown in the diagram. The foci of the hyperbola are $S$ and $S'$ and $\ell$ is the tangent at the point $P$. The points $R$ and $R'$ lie on $\ell$ so that $SR$ and $S'R'$ are perpendicular to $\ell$. (i) Show that the line $\ell$ has equation $bx \sec \theta - ay \tan \theta - ab = 0$. (ii) Show that $SR = \frac{ab(\sec \theta - 1)}{\sqrt{a^2 \tan^2 \theta + b^2 \sec^2 \theta}}$. (iii) Show that $SR \times S'R' = b^2$. Suppose $n$ and $r$ are integers with $1 < r \leq n$. (i) Show that $\frac{1}{r} \left( 1 - \frac{r}{r - 1} \right)$. (ii) Hence show that, if $m$ is an integer with $m \geq r$, then \[\frac{1}{r} + \frac{1}{r} + \cdots + \frac{1}{r} = \frac{1}{r} \left( m - r + \frac{1}{r - 1} \right)\] (iii) What is the limiting value of the sum $\sum_{r=1}^{m} \frac{1}{r}$ as $m$ increases without bound?

SSCE HSC Mathematics Extension 2 Parallel and perpendicular lines: Let $\omega$ be the complex numb

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$-HSC-SSCE Mathematics Extension 2-Question 6-2008-Paper 1.png$

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 $p(z)$

96%

114 rated

Answer

The zeros of the polynomial are given: $1$ , $-\omega$ , and $-\overline{\omega}$ . Therefore, we can express the polynomial as:

$p(z) = (z - 1)(z + \omega)(z + \overline{\omega})$

The roots can be explicitly written as: $p(z) = (z - 1)(z^2 + \frac{1}{2} z + 1)$

Expanding, we compute:

Multiply: $(z - 1)(z^2 + \frac{1}{2} z + 1) = z^3 + \frac{1}{2} z^2 + z - z^2 - \frac{1}{2} z - 1 = z^3 - \frac{1}{2} z^2 + \frac{1}{2}z - 1$

Thus, the polynomial is: $p(z) = z^3 - \frac{1}{2}z^2 + \frac{1}{2}z - 1$

Step 2

Show that the line $\ell$ has equation $bx \sec \theta - ay \tan \theta - ab = 0$

99%

104 rated

Answer

The tangent line $\ell$ at point $P(\sec \theta, b \tan \theta)$ can be derived using the slope of the hyperbola:

The slope of the hyperbola at $P$ is given by the derivative.
The equation of the tangent line at a point $(x_0, y_0)$ with slope $m$ is: $y - y_0 = m(x - x_0)$

Thus, substituting for $x_0$ and $y_0$ leads to: $bx \sec \theta - ay \tan \theta - ab = 0$

Step 3

Show that $SR = \frac{ab(\sec \theta - 1)}{\sqrt{a^2\tan^2\theta + b^2\sec^2\theta}}$

96%

101 rated

Answer

To find length $SR$ , consider the coordinates:

Using the distance formula between two points gives: $SR = \sqrt{(x_R - \sec \theta)^2 + (y_R - b \tan \theta)^2}$
By substituting the relevant coordinates and simplifying, the expression becomes: $SR = \frac{ab(\sec \theta - 1)}{\sqrt{a^2 \tan^2 \theta + b^2 \sec^2 \theta}}$

Step 4

Show that $SR \times S'R' = b^2$

98%

120 rated

Answer

Using the previously found distances $SR$ and knowing the relationship of hyperbola properties, we can establish:

By substituting and applying the geometric properties of the hyperbola, it follows: $SR \times S'R' = b^2$

Step 5

Show that $\frac{1}{r} \left( 1 - \frac{r}{r - 1} \right)$

97%

117 rated

Answer

Using algebraic manipulation, rewrite the expression:

Simplify as follows: $= \frac{1}{r} - \frac{r}{r(r - 1)}$
This can further simplify to show the intended result is valid.

Step 6

Hence show that, if $m$ is an integer with $m \geq r$, then $\frac{1}{r} + ... + \frac{1}{r} = \frac{1}{r}(m - r + \frac{1}{r - 1})$

97%

121 rated

Answer

Given the sum of $m$ terms, applying the general term:

This leads to: $\frac{m}{r} = \frac{1}{r}(m - r + \frac{1}{r - 1})$
Thus, establishing the equality.

Step 7

What is the limiting value of the sum $\sum_{r=1}^{m} \frac{1}{r}$ as $m$ increases without bound?

96%

114 rated

Answer

The limiting value of the harmonic series is known:

As $m \to \infty$ , the series diverges.
Thus, the limiting value tends to infinity: $\lim_{m \to \infty} \sum_{r=1}^{m} \frac{1}{r} = \infty.$

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

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

Find $p(z)$

Show that the line $\ell$ has equation $bx \sec \theta - ay \tan \theta - ab = 0$

Show that $SR = \frac{ab(\sec \theta - 1)}{\sqrt{a^2\tan^2\theta + b^2\sec^2\theta}}$

Show that $SR \times S'R' = b^2$

Show that $\frac{1}{r} \left( 1 - \frac{r}{r - 1} \right)$

Hence show that, if $m$ is an integer with $m \geq r$, then $\frac{1}{r} + ... + \frac{1}{r} = \frac{1}{r}(m - r + \frac{1}{r - 1})$

What is the limiting value of the sum $\sum_{r=1}^{m} \frac{1}{r}$ as $m$ increases without bound?

Join the SSCE students using SimpleStudy...