SCSA Mathematics Specialist Complex numbers

Given $z = r \text{ cis } \theta$ where $\tan \theta = \sqrt{2} \quad \therefore \frac{b}{3} = \sqrt{2}$ i.e. $b = 3\sqrt{2}$

$r^2 = 3^2 + b^2$

$r^2 = 9 + \left(3\sqrt{2}\right)^2 = 9 + 9(2) \quad \therefore r^2 = 27$

i.e. $r = \sqrt{27} = 3\sqrt{3}$

Given $w=x+i$

Let $w=r \operatorname{cis}(\theta)$

$w^7=r^7 \operatorname{cis}(7 \theta)$

$\arg (w)=\theta$

$=\tan ^{-1}\left(\frac{y}{x}\right), x \neq 0$

$=\tan ^{-1}\left(\frac{1}{x}\right)$ where $x \in R^{+}, x \neq 0$

$\arg \left(w^7\right)=7 \theta=7 \tan ^{-1}\left(\frac{1}{x}\right)$

Given $\operatorname{Re}\left(w^7\right)=0$, then $\arg \left(w^7\right)=(2 n+1) \frac{\pi}{2}$ where $n \in Z$

$7 \tan ^{-1}\left(\frac{1}{x}\right)=(2 n+1) \frac{\pi}{2}$

$\frac{1}{x}=\tan \left(\frac{(2 n+1) \pi}{14}\right)$

$x=\cot \left(\frac{(2 n+1) \pi}{14}\right)$

As $x \in R^{+}$, consider $\frac{(2 n+1) \pi}{14}$ for angles that lie in quadrant 1.

$n=0: x=\cot \left(\frac{\pi}{14}\right)$

$n=1: x=\cot \left(\frac{3 \pi}{14}\right)$

$n=2: x=\cot \left(\frac{5 \pi}{14}\right)$

The roots of $z^5 = 1$ are $z = \text{cis}\left(\frac{2k\pi}{5}\right)$ where $k \in Z$
Given $z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)$, the four roots of $z^4 + z^3 + z^2 + z + 1$ must be the four complex roots of the five 5th roots of unity, $z^5 = 1$.

By the conjugate root theorem, the two remaining roots of $P(z)$ must be a conjugate pair of roots of $z^5 = 1$.
One possible pair of roots is $\text{cis}\left(\frac{2\pi}{5}\right)$ and $\text{cis}\left(-\frac{2\pi}{5}\right)$

Determining a quadratic factor of $P(z)$
$\left(z - \text{cis}\left(\frac{2\pi}{5}\right)\right)\left(z - \text{cis}\left(-\frac{2\pi}{5}\right)\right)$
$= z^2 - \left(\text{cis}\left(\frac{2\pi}{5}\right) + \text{cis}\left(-\frac{2\pi}{5}\right)\right)z + \text{cis}\left(\frac{2\pi}{5}\right)\text{cis}\left(-\frac{2\pi}{5}\right)$
$= z^2 - \left(\cos\left(\frac{2\pi}{5}\right) + i\sin\left(\frac{2\pi}{5}\right) + \cos\left(-\frac{2\pi}{5}\right) + i\sin\left(-\frac{2\pi}{5}\right)\right)z + \text{cis}(0)$
$= z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1$

$P(z) = (z-2)\left(z^2 - 2\cos\left(\frac{2\pi}{5}\right)z + 1\right)$
$= z^3 - 2\left(\cos\left(\frac{2\pi}{5}\right) + 1\right)z^2 + \left(4\cos\left(\frac{2\pi}{5}\right) + 1\right)z - 2$

$|z^4| = \sqrt{2^2+(2\sqrt{3})^2} = \sqrt{4+12} = 4$
$Arg(z^4) = \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$

$\therefore \text{Solve } z^4 = 4\text{cis}\left(-\frac{\pi}{3}\right)$

$\therefore z = 4^{\frac{1}{4}}\text{cis}\left(\frac{-\pi}{12}+k\left(\frac{\pi}{2}\right)\right) \quad k = 0, 1, 2, 3 \quad ... (1)$

Roots are:
$z_0 = \sqrt{2}\text{cis}\left(-\frac{\pi}{12}\right)$
$z_1 = \sqrt{2}\text{cis}\left(\frac{5\pi}{12}\right)$
$z_2 = \sqrt{2}\text{cis}\left(\frac{11\pi}{12}\right)$
$z_3 = \sqrt{2}\text{cis}\left(-\frac{7\pi}{12}\right) \quad \text{Note: } z_3 = \sqrt{2}\text{cis}\left(\frac{17\pi}{12}\right) \text{ does not satisfy } -\pi < \theta \le \pi.$

$P(1+i) = (1+i)^2 - 2(1+i) + 2 = 2i - 2 - 2i + 2 = 0$
$\therefore P(1+i) = 0$
Hence $(z-1-i)$ is a factor of $P(z)$.

$z = 1-i$ is also a root of $P(z)$ and $R(z)$.

$R(z) = z^4 - 6z^3 + 17z^2 - 22z + 14 = (z^2 - 2z + 2)(z^2 + az + b)$
$\therefore 2b = 14 \quad \implies b = 7$
$-6z^3 = z^2(az) - 2z(z^2) \quad \implies a = -4$

Solve $R(z) = (z^2 - 2z + 2)(z^2 - 4z + 7) = 0:$

$(z^2 - 2z + 2) = 0 \implies \therefore z = 1 \pm i$
and

$P(z) = 2z^4 + az^3 + 6z^2 + az + b$ where $a, b \in Z^+$
Given $z = -i$ is a root of $P(z)$, then $P(-i) = 0$
$\therefore 2(-i)^4 + a(-i)^3 + 6(-i)^2 + a(-i) + b = 0$
$2 + ai - 6 - ai + b = 0$
$-4 + b = 0$
$b = 4$
$\therefore P(z) = 2z^4 + az^3 + 6z^2 + az + 4$

Given that the coefficients of the polynomial are real, another root is $z = i$, another factor of $P(z)$ is $(z - i)$.
$(z - i)(z + i) = (z^2 + 1)$ is a factor of $P(z)$

By inspection,
$P(z) = (z^2 + 1)(2z^2 + az + 4)$

Given all roots of $P(z)$ have an imaginary component, $2z^2 + az + 4$ must have only complex roots.
For complex roots, $b^2 - 4ac < 0$
$a^2 - 4 \times 2 \times 4 < 0$
$a < \sqrt{32}$
So $a = 1, 2, 3, 4$ or $5$ and $b = 4$

RTP
$|z - w|^2 = |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$

LHS $= |z - w|^2$
$= |(a + bi) - (c + di)|^2$
$= |(a - c) + (b - d)i|^2$
$= (a - c)^2 + (b - d)^2$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$

RHS $= |z|^2 + |w|^2 - 2\text{Re}(z \overline{w})$
$= |a + bi|^2 + |c + di|^2 \dots$
$\dots - 2\text{Re}((a + bi)(c - di))$
$= a^2 + b^2 + c^2 + d^2 \dots$
$\dots - 2\text{Re}((ac + bd) + (bc - ad)i)$
$= a^2 + b^2 + c^2 + d^2 - 2(ac + bd)$
$= a^2 + b^2 + c^2 + d^2 - 2ac - 2bd$
$= a^2 - 2ac + c^2 + b^2 - 2bd + d^2$
$= \text{LHS}$