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

The powers of $i$ cycle every four:

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

Thus:

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

Combining:

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

Final result:

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

To find the angle $\theta$ between two vectors, we use the formula:

$\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{||\mathbf{a}|| \, ||\mathbf{b}||}$

Calculating the dot product:

$\mathbf{a} \cdot \mathbf{b} = 2(-3) + 4(1) = -6 + 4 = -2$

Calculating the magnitudes:

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

Substituting into the cosine formula:

$\cos(\theta) = \frac{-2}{(2\sqrt{5})(\sqrt{10})} = \frac{-2}{2\sqrt{50}} = \frac{-1}{\sqrt{50}} = -\frac{1}{5\sqrt{2}}$

Using a calculator, find:

$\theta = \arccos\left(-\frac{1}{5\sqrt{2}}\right)$

Calculating this yields an angle of approximately $118.7$ degrees (correct to 1 decimal place).

We can express $-i$ in polar form as:

$-i = e^{-i\frac{\pi}{2}}$

To find the square roots, we need:

$\sqrt{-i} = e^{-i\frac{\pi}{4}} = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$

Also, the other root will be:

$\sqrt{-i} = e^{-i\frac{\pi}{4} + i\pi} = e^{i\frac{3\pi}{4}} = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$

Hence, the two square roots are:

$x + iy = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} \quad \text{and} \quad -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$

Using the quadratic formula:

$z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Substituting $a = 1$, $b = 2$, and $c = 1 + i$:

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

$= \frac{-2 \pm \sqrt{4 - (4 + 4i)}}{2} = \frac{-2 \pm \sqrt{-4i}}{2}$

This requires evaluating $\rac{-2 \pm 2\sqrt{2}e^{-i\frac{\pi}{4}}}{2}$ resulting in:

$z = -1 \pm \sqrt{2}e^{-i\frac{\pi}{4}}$

Given $z = 5 + i$ and $w = 2 - 4i$, we can find:

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

To simplify, multiply by the conjugate of the denominator:

$= \frac{(5 + i)(2 + 4i)}{(2 - 4i)(2 + 4i)} = \frac{10 + 20i + 2i + 4i^{2}}{4 + 16}$ $= \frac{10 + 22i - 4}{20} = \frac{6 + 22i}{20} = \frac{3}{10} + i\frac{11}{10}$

Thus, the answer is:

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

We set:

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

Multiplying through by the denominator:

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

Equating coefficients:

For $x^{2}$: $3 = A + B$ For $x^{1}$: $0 = A - 2B + C$ For $x^{0}$: $-5 = A - 2C$

Solving these will provide the values of $A$, $B$, and $C$.