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

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

$\frac{-2 + 3i}{2 + i} \cdot \frac{2 - i}{2 - i} = \frac{(-2 + 3i)(2 - i)}{(2 + i)(2 - i)}.$ The denominator simplifies as: $(2 + i)(2 - i) = 4 + 1 = 5.$ The numerator becomes: $-4 + 2i + 6i - 3 = -7 + 8i.$ Therefore, $\frac{-2 + 3i}{2 + i} = \frac{-7 + 8i}{5} = -\frac{7}{5} + i\frac{8}{5}.$ Thus, $a = -\frac{7}{5}$ and $b = \frac{8}{5}$.

To mark the point $R$ representing $iz$, first calculate $iz$: if $z = a + bi$, then $iz = i(a + bi) = -b + ai$. It should be marked on the Argand diagram at coordinates $(-b, a)$.

For the point $S$ representing $w$, the coordinates will depend on the specific value of $w = c + di$. It should be marked on the Argand diagram at $(c, d)$ where $c$ and $d$ are the real and imaginary parts of $w$, respectively.

For the point $T$ representing $z + w$, sum the corresponding coordinates: if $z = a + bi$ and $w = c + di$, then $z + w = (a + c) + (b + d)i$. Mark point $T$ at $(a + c, b + d)$ in the Argand diagram.

The inequality $|z - 1| \\leq 2$ represents a circle with center at $(1, 0)$ and radius $2$. The argument conditions give a sector in the complex plane starting from the point $1 - 2i$ to the point $1 + 2i$ at an angle of $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ from the positive real axis. Sketch this sector within the circle.

To find the 5th roots of $-1$, express $-1$ in polar form: $-1 = 1 e^{i\pi}.$ The 5th roots can be found using De Moivre's theorem: $z_k = 1^{\frac{1}{5}} e^{i\frac{\pi + 2k\pi}{5}}\text{ for } k = 0, 1, 2, 3, 4.$ This results in: $z_0 = e^{i\frac{\pi}{5}}, z_1 = e^{i\frac{\pi}{5} + \frac{2\pi}{5} }, z_2 = e^{i\frac{\pi}{5} + \frac{4\pi}{5}}, z_3 = e^{i\frac{\pi}{5} + \frac{6\pi}{5}}, z_4 = e^{i\frac{\pi}{5} + \frac{8\pi}{5}}.$

On the Argand diagram, plot the points found for the roots $z_k$ with $k = 0, 1, 2, 3, 4$. The points will be evenly distributed around the unit circle, at angles of:

• $\frac{\pi}{5}$
• $\frac{3\pi}{5}$
• $\pi$
• $\frac{7\pi}{5}$
• $\frac{9\pi}{5}$. Mark each point accordingly on the unit circle.

To find the square roots of $3 + 4i$, express $z = x + yi$ such that:

$(x + yi)^2 = 3 + 4i.$ This expands to: $x^2 - y^2 + 2xyi = 3 + 4i.$ Thus, we can form the system of equations:

1. $x^2 - y^2 = 3$
2. $2xy = 4$. Solving Equation 2 gives $xy = 2$, from which $y = \frac{2}{x}$. Substituting this into Equation 1 provides: $x^2 - \left(\frac{2}{x}\right)^2 = 3,$ leading to: $x^4 - 3x^2 - 4 = 0.$ Letting $u = x^2$, we can re-write this as: $u^2 - 3u - 4 = 0.$ Solving using the quadratic formula: $u = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}.$ Thus, $u = 4 ext{ or } -1.$ Hence, $x = 2$ or $y = 0$. The roots will then be: $z = 2 + i$ and $z = -2 - i$.

To solve the equation $z^2 + iz - 1 = 0$, we can apply the quadratic formula: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = i$, and $c = -1$. This gives: $z = \frac{-i \pm \sqrt{i^2 + 4}}{2},$ leading to: $z = \frac{-i \pm \sqrt{3}}{2}.$ The solutions are thus: $z = \frac{-i + \sqrt{3}}{2} \text{ and } z = \frac{-i - \sqrt{3}}{2}.$