Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet. On the diagram provided in the writing booklet, indicate the locations of all of the fourth roots of the complex number a + ib. (b) Use an appropriate substitution to evaluate \[ \int_{\sqrt{3}}^{\sqrt{10}} \frac{x^3}{\sqrt{x^2 - 9}} \, dx. \]

SSCE HSC Mathematics Extension 2 Complex numbers: Use the Question 13 Writing Book

Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2021 - Paper 1

Question 13

Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet. On the diag... show full transcript

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2021 - Paper 1

Step 1

The location of all of the fourth roots of the complex number a + ib

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Answer

To indicate the locations of the fourth roots of the complex number $a + ib$ , we first need to express it in polar form. The modulus of the complex number is given by $r = \sqrt{a^2 + b^2}$ . The argument (angle) is $\theta = \tan^{-1}(\frac{b}{a})$ . The fourth roots can be computed using the formula:

$z_k = r^{1/4} \left( \cos\left( \frac{\theta + 2k\pi}{4} \right) + i \sin\left( \frac{\theta + 2k\pi}{4} \right) \right), \, k = 0, 1, 2, 3$

Indicate the roots on the provided diagram by plotting the points corresponding to these values.

Step 2

Use an appropriate substitution to evaluate

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Answer

To evaluate the integral:

$\int_{\sqrt{3}}^{\sqrt{10}} \frac{x^3}{\sqrt{x^2 - 9}} \, dx,$

we can use the substitution $u = x^2 - 9$ . Then, we find that $du = 2x \, dx$ , or $dx = \frac{du}{2x}$ . The limits change as follows:

When $x = \sqrt{3}$ , $u = 3 - 9 = -6$
When $x = \sqrt{10}$ , $u = 10 - 9 = 1$

Inserting these values into the integral, we express $x$ in terms of $u$ , giving:

$\int_{-6}^{1} \frac{(u + 9)\sqrt{u + 9}}{\sqrt{u}} \frac{du}{2\sqrt{u + 9}} = \frac{1}{2} \int_{-6}^{1} (u + 9) \frac{u^{1/2}}{u^{1/2}} \, du$

From here, we will simplify and carry out the integration, being careful with the algebra to reach the final answer.

Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2021 - Paper 1

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) The location of the complex number a + ib is shown on the diagram on page 1 of the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2021 - Paper 1

The location of all of the fourth roots of the complex number a + ib

Use an appropriate substitution to evaluate

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