Write $i^3$ in the form $a + ib$ where $a$ and $b$ are real - HSC - SSCE Mathematics Extension 2 - Question 2 - 2009 - Paper 1

The complex number is already in the form $a + ib$, where:

a = -2, b = -3.

So, we can express it simply as:

$-2 - 3i$

If $z = a + bi$, then:

$R = i(a + bi) = -b + ai$

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

If $w = c + di$, then simply mark the point at $(c, d)$ on the diagram.

To find $T$, we compute:

$T = (a + bi) + (c + di) = (a + c) + (b + d)i$

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

The inequality $|z - 1| \leq 2$ represents a circle centered at $(1, 0)$ with radius 2. The argument inequality defines a sector of this circle from angle $-\frac{\pi}{4}$ to $\frac{\pi}{4}$, which corresponds to the region in the first quadrant and a portion of the fourth quadrant. Overall, shade this sector.

The modulus of $-1$ is $1$, and the argument is $\\pi$. The 5th roots can be found as:

$r_k = 1^{1/5} \quad \text{and} \quad \theta_k = \frac{\pi + 2k\pi}{5}, \; k = 0, 1, 2, 3, 4$

This gives the roots:

• For $k=0$: $\theta_0 = \frac{\pi}{5}$.
• For $k=1$: $\theta_1 = \frac{\pi + 2\pi}{5} = \frac{3\pi}{5}$.
• For $k=2$: $\theta_2 = \frac{\pi + 4\pi}{5} = \frac{5\pi}{5} = \pi$.
• For $k=3$: $\theta_3 = \frac{\pi + 6\pi}{5} = \frac{7\pi}{5}$.
• For $k=4$: $\theta_4 = \frac{\pi + 8\pi}{5} = \frac{9\pi}{5}$.

Plot the computed roots on an Argand diagram using their polar forms:

$e^{i\frac{\pi}{5}}, e^{i\frac{3\pi}{5}}, e^{i\pi}, e^{i\frac{7\pi}{5}}, e^{i\frac{9\pi}{5}}$

The points will be equally spaced on the unit circle, with angles as calculated.

To find the square roots, let $z = x + yi$. Then:

$(x + yi)^2 = 3 + 4i$

Expanding gives:

$x^2 - y^2 + 2xyi = 3 + 4i$

Comparing real and imaginary parts:

1. $x^2 - y^2 = 3$
2. $2xy = 4\implies xy = 2$

From $xy = 2$, we can express $y = \frac{2}{x}$ and substitute into the first equation:

$x^2 - \left(\frac{2}{x}\right)^2 = 3$

Clearing the fraction results in:

$x^4 - 3x^2 - 4 = 0$

Letting $u = x^2$, we get:

$u^2 - 3u - 4 = 0$

Applying the quadratic formula yields:

$u = \frac{3 \pm \sqrt{(3)^2 + 4 \cdot 4}}{2} = \frac{3 \pm 5}{2}$

Thus, $u = 4$ or $u = -1$. Since $u = x^2$, we discard $u = -1$:

1. $x^2 = 4 \implies x = 2 \; or \; -2$.

Using $y = \frac{2}{x}$ leads us to two pairs:

• For $x = 2$, $y = 1$ giving the root $2 + i$.
• For $x = -2$, $y = -1$ giving the root $-2 - i$.

Thus, the square roots of $3 + 4i$ are: $2 + i, -2 - i$

Using the quadratic formula, we have:

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

Where:

• $a = 1$,
• $b = i$,
• $c = -1$. Thus:

$z = \frac{-i \pm \sqrt{i^2 - 4(1)(-1)}}{2} = \frac{-i \pm \sqrt{-1 + 4}}{2} = \frac{-i \pm \sqrt{3}}{2}$

Therefore, the solutions are:

$z = \frac{-i + \sqrt{3}}{2} \quad \text{and} \quad z = \frac{-i - \sqrt{3}}{2}$