Write $i^n$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

To write $\frac{-2 + 3i}{2 + i}$ in the form $a + ib$, we multiply the numerator and denominator by the conjugate of the denominator:

$\frac{(-2 + 3i)(2 - i)}{(2 + i)(2 - i)}$

Calculating the denominator: $(2 + i)(2 - i) = 4 + 1 = 5$

Calculating the numerator: $(-2 + 3i)(2 - i) = -4 + 2i + 6i + 3 = -1 + 8i$

Thus, we have: $\frac{-1 + 8i}{5} = -\frac{1}{5} + \frac{8}{5}i$

So, $a = -\frac{1}{5}$ and $b = \frac{8}{5}$.

To find the point $R$, we calculate $iz$ where $z$ is the complex number. If $z = a + bi$, then: $R = iz = i(a + bi) = -b + ai$

Mark the point $R$ at the coordinates $(-b, a)$ on the Argand diagram.

The point $S$ represents the complex number $w$. If $w = c + di$, mark the point $S$ on the Argand diagram at the coordinates $(c, d)$.

The point $T$ denotes $z + w$. If $z = a + bi$ and $w = c + di$, then: $T = (a + c) + (b + d)i$

Mark the point $T$ on the Argand diagram at the coordinates $(a + c, b + d)$.

The inequality $|z - 1| \leq 2$ represents a circle centered at $(1, 0)$ with radius 2. The inequality $-\frac{\pi}{4} \leq arg(z - 1) \leq \frac{\pi}{4}$ restricts $z$ to the sector between the angles $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ originating from the point $(1, 0)$. Sketch the circle and shade the area that also falls within these angles.

The modulus of $-1$ is 1 and the argument is $\pi + 2k\pi$ for $k \in \mathbb{Z}$.

To find the 5th roots, we use the formula: $z_k = r^{1/5} \left( \cos \left( \frac{\pi + 2k\pi}{5} \right) + i \sin \left( \frac{\pi + 2k\pi}{5} \right) \right), k = 0, 1, 2, 3, 4$

Calculating:

• For $k = 0$: $z_0 = \cos \left( \frac{\pi}{5} \right) + i \sin \left( \frac{\pi}{5} \right)$
• For $k = 1$: $z_1 = \cos \left( \frac{\pi + 2\pi}{5} \right) + i \sin \left( \frac{\pi + 2\pi}{5} \right)$
• Continue for $k = 2, 3, 4$.

On the Argand diagram, plot the roots found previously:

• Each root will be at an angle of $\frac{\pi + 2k\pi}{5}$ from the positive real axis with a radius of 1.
• Ensure to evenly distribute the points around the unit circle.

To find the square roots of $3 + 4i$, we assume: $z = x + yi$ Then: $z^2 = 3 + 4i$ Expanding gives: $(x^2 - y^2) + 2xyi = 3 + 4i$ Equating real and imaginary parts:

• $x^2 - y^2 = 3$
• $2xy = 4 \Rightarrow xy = 2$

Solve these equations to find $x$ and $y$.

Rearranging gives: $z^2 + iz - 1 - i = 0$ $z^2 + iz - (1 + i) = 0$ Using the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ With $a = 1$, $b = i$, and $c = -(1 + i)$, calculate the roots.