Question 18
Consider the function $P(z)=2z^{4}+az^{3}+6z^{2}+az+b$ where $a,b\in\mathbb{Z}^{+}$
One of the roots of $P(z)$ is $z=-i$
Determine the possible value/s for $a$ and $b$ such that all remaining roots of $P(z)$ have an imaginary component.
Two roots are fixed, and a moves the other two
Watch the roots of $P(z)$ on an Argand diagram as $a$ increases. The orange roots turn white once they land on the real axis.
The orange roots always multiply to 2, which is why they stay on the circle $|z|=\sqrt{2}$ until they become real.
Find every a that keeps all the roots complex
$a$ has to be a positive integer. Step through the values and watch the roots of $2z^{2}+az+4$.
Work through the solution one mark at a time
Try each step yourself before you reveal it. There is no calculator, so the powers of $i$ need to be right.
$z=-i$ is a root, so $P(-i)=0$. Use $(-i)^{2}=-1$, $(-i)^{3}=i$ and $(-i)^{4}=1$.
$$2(-i)^{4}+a(-i)^{3}+6(-i)^{2}+a(-i)+b=0$$
$$2+ai-6-ai+b=0\quad\Rightarrow\quad b=4$$
Marker: correctly applies the factor theorem to determine $b$
The coefficients of $P(z)$ are real, so by the conjugate root theorem $z=i$ is also a root. That gives you another factor.
$$(z-i)\ \text{is a factor of}\ P(z)$$
Marker: correctly uses the conjugate root of the given root to identify another factor of $P(z)$
Multiply the two factors together.
$$(z-i)(z+i)=z^{2}+1$$
This mark can be implied by your later working.
Marker: correctly identifies that $\left(z^{2}+1\right)$ is a factor of $P(z)$
Find the other quadratic factor. You need $2z^{2}$ to make $2z^{4}$, and $4$ to make the constant 4. Then check the middle terms.
$$P(z)=\left(z^{2}+1\right)\left(2z^{2}+az+4\right)$$
Polynomial division also works. You keep follow-through marks from here if you slipped earlier.
Marker: determines the remaining quadratic factor in terms of $a$
“All remaining roots have an imaginary component” means $2z^{2}+az+4$ must not have real roots. So its discriminant has to be negative.
$$a^{2}-4(2)(4)<0\quad\Rightarrow\quad a^{2}<32\quad\Rightarrow\quad a<\sqrt{32}$$
You only need the positive square root, because $a$ is a positive integer.
Marker: applies the complex root requirement to the remaining quadratic factor
$\sqrt{32}$ is between 5 and 6, because $5^{2}=25$ and $6^{2}=36$. $a$ is a positive integer, so list every value.
$a=1,\ 2,\ 3,\ 4$ or $5$, and $b=4$
Make sure you give $b$ as well. The question asks for the values of both $a$ and $b$.
Marker: determines the possible values for $a$ given $a,b\in\mathbb{Z}^{+}$
Each phrase of the question gave you something
There are two unknowns but only one root to use. The substitution only gives you $b$, so you have to factorise $P(z)$ and turn a sentence about the roots into an inequality for $a$.
I would check that you listed every value of $a$. The QCAA like to add conditions such as $a,b\in\mathbb{Z}^{+}$, so the answer is a short list rather than a single number.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.