The complex numbers $ z = 2 e^{i \frac{\pi}{2}} $ and $ w = 6 e^{i \frac{\pi}{6}} $ are given - HSC - SSCE Mathematics Extension 2 - Question 11 - 2021 - Paper 1

To compute the sum:

$\sum_{n=1}^{5} (i)^{n} = i + i^{2} + i^{3} + i^{4} + i^{5}$

Using the powers of $i$:

• $i^{1} = i$
• $i^{2} = -1$
• $i^{3} = -i$
• $i^{4} = 1$
• $i^{5} = i$

Then, summing:

$= i - 1 - i + 1 + i = i + (-1 + 1) = i$

So, the result is:

$\sum_{n=1}^{5} (i)^{n} = i$

To find the angle $\theta$ between the vectors $\mathbf{q} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$:

1. Calculate the dot product: $\mathbf{q} \cdot \mathbf{b} = 2(-3) + 4(1) = -6 + 4 = -2$

2. Find the magnitudes:

   • For $\mathbf{q}$: $|\mathbf{q}| = \sqrt{2^{2} + 4^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$
   • For $\mathbf{b}$: $|\mathbf{b}| = \sqrt{(-3)^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10}$

3. Use the dot product formula: $\cos \theta = \frac{\mathbf{q} \cdot \mathbf{b}}{|\mathbf{q}| |\mathbf{b}|} = \frac{-2}{(2\sqrt{5})(\sqrt{10})} = \frac{-2}{2\sqrt{50}} = \frac{-1}{\sqrt{50}} = -\frac{1}{5\sqrt{2}}$

4. Calculate $\theta$: $\theta = \cos^{-1}\left( -\frac{1}{5\sqrt{2}} \right) \approx 118.1^{\circ}$

Thus, the angle is approximately:

$\theta \approx 118.1^{\circ}$

Let $z = x + iy$, where $x$ and $y$ are real numbers. We need to find:

$z^{2} = -i$

Expanding gives:

$(x + iy)^{2} = x^{2} + 2xy i - y^{2} = -i$

From this, separate real and imaginary parts:

1. Real part: $x^{2} - y^{2} = 0$ This implies $x^{2} = y^{2} \Rightarrow y = \pm x$
2. Imaginary part: $2xy = -1 \Rightarrow xy = -\frac{1}{2}$

Substituting $y = x$ gives: $x^{2} = -\frac{1}{2}$ which has no real solutions. Using $y = -x$ we have: $-x^{2} = -\frac{1}{2} \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \mp \frac{1}{\sqrt{2}}$

Thus, the square roots are:

$\sqrt{-i} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} i ext{ and } -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i$.

From part (i), we have:

1. Let $z^{2} + 2z + (1 + i) = 0$
2. Use the quadratic formula:

$z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^{2} - 4(1)(1+i)}}{2(1)}$

Computing the discriminant: $b^{2} - 4ac = 4 - 4(1 + i) = 4 - 4 - 4i = -4i$

So, we have: $z = \frac{-2 \pm \sqrt{-4i}}{2} = -1 \pm \frac{\sqrt{4} \cdot e^{i(\frac{\pi}{2})}}{2}$

Thus: $z = -1 \pm (1 \cdot e^{i(\frac{\pi}{6})}) = -1 + e^{i(\frac{\pi}{6})} ext{ and } -1 - e^{i(\frac{\pi}{6})}.$

Given: $z = 5 + i ext{ and } w = 2 - 4i$

Calculate: $\frac{z}{w} = \frac{5 + i}{2 - 4i}$

Multiply by the conjugate of the denominator: $\frac{(5 + i)(2 + 4i)}{(2 - 4i)(2 + 4i)} = \frac{10 + 20i + 2i - 4}{4 + 16} = \frac{6 + 22i}{20} = \frac{3}{10} + \frac{11}{10}i$

Thus, the result in Cartesian form is:

$\frac{z}{w} = \frac{3}{10} + \frac{11}{10}i$

Let: $\frac{3x^{2} - 5}{(x - 2)(x^{2} + x + 1)} = \frac{A}{x - 2} + \frac{Bx + C}{x^{2} + x + 1}$

Multiply through by the common denominator: $3x^{2} - 5 = A(x^{2} + x + 1) + (Bx + C)(x - 2)$

Expanding the right hand side: $3x^{2} - 5 = Ax^{2} + Ax + A + Bx^{2} - 2Bx + Cx - 2C$

Combine like terms: $(A + B)x^{2} + (A - 2B + C)x + (A - 2C)$

Set coefficients equal:

1. $A + B = 3$
2. $A - 2B + C = 0$
3. $A - 2C = -5$

Solving these gives: After substitution and solving: $A = 7, B = -4, C = 7$

Thus, the partial fraction decomposition is: $\frac{3x^{2} - 5}{(x - 2)(x^{2} + x + 1)} = \frac{7}{x - 2} + \frac{-4x + 7}{x^{2} + x + 1}$