QCAA Specialist Mathematics Alternative Sequence Further complex numbers

Method 1
Prove $\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}$, given $z_1 = a + bi, z_2 = c + di, z_2 \neq 0$
$\text{LHS} = \left| \frac{a + bi}{c + di} \right|$
$= \left| \frac{(a + bi)}{(c + di)} \cdot \frac{(c - di)}{(c - di)} \right|$
$= \left| \frac{ac - adi + bci - bdi^2}{c^2 + d^2} \right|$
$= \left| \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} \right|$
$= \sqrt{\frac{(ac + bd)^2 + (bc - ad)^2}{(c^2 + d^2)^2}}$
$= \sqrt{\frac{(ac)^2 + 2abcd + (bd)^2 + (bc)^2 - 2abcd + (ad)^2}{(c^2 + d^2)^2}}$
$= \sqrt{\frac{(ac)^2 + (bd)^2 + (bc)^2 + (ad)^2}{(c^2 + d^2)^2}}$
$= \sqrt{\frac{a^2(c^2 + d^2) + b^2(c^2 + d^2)}{(c^2 + d^2)^2}}$
$= \sqrt{\frac{(a^2 + b^2)(c^2 + d^2)}{(c^2 + d^2)^2}}$
$= \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}$
$= \frac{|z_1|}{|z_2|}$
$= \text{RHS} \quad \text{QED}$

$P(z)=2z^4+az^3+6z^2+az+b$ where $a,b\in\mathbb{Z}^+$
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 $b=4$

$P(z) = z^3 + (i - a)z^2 - 2biz + 3i$
Using the remainder theorem
$P(2i) = a - bi$

By substitution
$P(2i) = (2i)^3 + (i - a)(2i)^2 - 2bi(2i) + 3i$
$= -8i - 4(i - a) + 4b + 3i$
$= -8i - 4i + 4a + 4b + 3i$
$= 4a + 4b - 9i$

Equating parts:
$a = 4a + 4b$
$-b = -9$

So $a = -12$ and $b = 9$
$\therefore P(z) = z^3 + (i + 12)z^2 - 18iz + 3i$

$P(a - bi) = P(-12 - 9i) \approx 1566 - 285i$

Since $P(-12 - 9i) \neq 0$, it is not reasonable that $(z - (a - bi))$ is a factor of $P(z)$.