(a) Express \( \frac{4+3i}{2-i} \) in the form \( x + iy \), where \( x \) and \( y \) are real - HSC - SSCE Mathematics Extension 2 - Question 11 - 2015 - Paper 1

The modulus of a complex number ( z = a + bi ) is given by ( |z| = \sqrt{a^2 + b^2} ).

For ( z = -\sqrt{3} + i ):

$$|z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2.$$

The argument of a complex number ( z = a + bi ) is given by ( \text{arg}(z) = \tan^{-1}\left( \frac{b}{a} \right) ).

For ( z = -\sqrt{3} + i ):

$$\text{arg}(z) = \tan^{-1}\left( \frac{1}{-\sqrt{3}} \right).$$

This point is in the second quadrant, so we have:

$$\text{arg}(z) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.$$

To find the argument of ( \frac{z}{w} ), we apply the property that ( \text{arg}\left( \frac{z}{w} \right) = \text{arg}(z) - \text{arg}(w) ).

For the complex number ( w = 3 \left( \cos \frac{\pi}{7} + i \sin \frac{\pi}{7} \right) ), we have:

$$\text{arg}(w) = \frac{\pi}{7}.$$

Thus:

$$\text{arg}\left( \frac{z}{w} \right) = \frac{5\pi}{6} - \frac{\pi}{7}.$$

Calculating this gives:

$$\text{arg}\left( \frac{z}{w} \right) = \frac{35\pi - 6\pi}{42} = \frac{29\pi}{42}.$$

To find ( A, B, ) and ( C ), we multiply both sides by ( (x^2 + 2)x^2 ):

$$1 = A x^2 + (Bx + C)(x^2 + 2).$$

Expanding gives:

$$1 = Ax^2 + Bx^3 + (2B + C)x^2 + 2C.$$

Comparing coefficients, we set up the equations:

• For ( x^3 ): ( B = 0 )
• For ( x^2 ): ( A + 2C = 0 )
• For the constant term: ( 2C = 1 )

From ( 2C = 1 ), we find ( C = \frac{1}{2} ). Substituting into the second equation gives ( A + 1 = 0 ), thus ( A = -1 ). Therefore,:

$$A = -1, B = 0, C = \frac{1}{2}.$$

The equation represents an ellipse centered at the origin. The semi-major axis ( a = 5 ) and semi-minor axis ( b = 4 ). The foci are located at a distance ( c = \sqrt{a^2 - b^2} ) from the center.

Calculating gives:

$$c = \sqrt{25 - 16} = \sqrt{9} = 3.$$

Thus, the foci are at ( (\pm 3, 0) ). The sketch should indicate the ellipse and the foci points at ( (3, 0) ) and ( (-3, 0) ).

First, rewrite the curve Equation as:

$$\frac{dy}{dx} = \frac{d}{dx}(x + x^3).$$

Finding the derivative gives:

$$\frac{dy}{dx} = 1 + 3x^2.$$

Substituting in the point ( x = 2 ):

$$\frac{dy}{dx} = 1 + 3(2^2) = 1 + 12 = 13.$$

Using the definitions of ( \cot \theta ) and ( \csc \theta ):

$$\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \csc \theta = \frac{1}{\sin \theta},$$

we have:

$$\cot \theta + \csc \theta = \frac{\cos \theta + 1}{\sin \theta}.$$

Recall the double angle formulas, we find:

$$$$

We know that:

$$d\theta = - \log|\sin \theta| + C = -\ln |\sin \theta| + C,$$

therefore,

$$\int \left( \cot \theta + \csc \theta \right) d\theta = \ln |\sin \theta| + C.$$