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

For the complex number ( z = -\sqrt{3} + i ), its modulus is given by:

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

The argument of the complex number ( z = -\sqrt{3} + i ) is:

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

\nThis point is located in the second quadrant, so:

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

We need to find the argument of the quotient of two complex numbers. The argument of a quotient is given as:

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

\nWe previously found ( \text{arg}(z) = \frac{5\pi}{6} ). Now we need to find ( \text{arg}(w) ).

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

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

\nThus:

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

Computing this gives:

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

To find the values of ( A, B, ) and ( C ), we equate the expression:

$$1 = A + \frac{Bx + C}{x^2} (x^2 + 2).$$

\nMultiplying through by ( x^2(x^2 + 2) ) gives:

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

Setting coefficients equal, we can derive the equations to solve for ( A, B, ) and ( C ). After simplification and comparison, we find: ( A = 1, B = 0, C = -2. )

The equation ( \frac{x^2}{25} + \frac{y^2}{16} = 1 ) represents an ellipse with:

• Semi-major axis ( a = 5 ) (along the x-axis)
• Semi-minor axis ( b = 4 ) (along the y-axis)

To find the foci, we use the relationship:

$$c = \sqrt{a^2 - b^2} = \sqrt{25 - 16} = \sqrt{9} = 3.$$

\nThe foci are then located at: ( (\pm c, 0) = (\pm 3, 0). ) \nThus, we mark the foci at the points ( (3, 0) ) and ( (-3, 0) ) on our sketch.

To find ( \frac{dy}{dx} ), we can use implicit differentiation on the equation ( x + x^3 + 2 = 0 ):

$$1 + 3x^2 \frac{dx}{dy} = 0 \Rightarrow \frac{dy}{dx} = -\frac{1}{3x^2}.$$

\nSubstituting ( x = 2 ):

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

Starting with the left-hand side:

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

Applying the angle half-formula:

$$\cot\left(\frac{\theta}{2}\right) = \frac{\cos\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} = \frac{2\cos^{2}\left(\frac{\theta}{2}\right)}{1 - \cos \theta}.$$

\nThus, with further simplifications, we confirm that:

$$\cot \theta + \csc \theta = \cot\left(\frac{\theta}{2}\right).$$

To solve the integral:

$$\int (\cot \theta + \csc \theta) d\theta = \int \cot \theta d\theta + \int \csc \theta d\theta.$$

Using known results for each term:

$$\int \cot \theta d\theta = \ln |\sin \theta| + C \quad \text{and} \quad \int \csc \theta d\theta = -\ln |\csc \theta + \cot \theta| + C.$$

\nThus, the complete integral becomes:

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