Question 2
When the polynomial $P(z)=z^3-iz^2$ $-\,z-i$ is divided by $z-i$, the remainder is
That’s it. Step through the hints below to see why. In 2021, 64.0% of Queensland students chose (A).
Not quite. $0$ comes from taking $i^3=i$. In fact $i^3=i^2\times i=-i$. With $i^3=i$ the terms become $i+i-i-i=0$. A remainder of $0$ would also mean $z-i$ is a factor, and it isn’t. In 2021, 18.7% of Queensland students chose (B).
Not quite. $2i$ is $P(-i)$: the terms are $i+i+i-i=2i$. That is the remainder when you divide by $z+i$. For $z-i$ you substitute the value that makes it zero, $z=i$. In 2021, 13.8% of Queensland students chose (C).
Not quite. $4i$ is what you get if you substitute $z=-i$ and also drop the minus sign on the constant term: $i+i+i+i=4i$. Substitute $z=i$, and keep every sign. In 2021, 2.8% of Queensland students chose (D).
Every term of $P(z)$ is purely imaginary at $z=\pm i$, so each one is an arrow up or down the imaginary scale. The orange arrow is their total.
$z=i$ is the value that makes the divisor $z-i$ zero, so $P(i)=-2i$ is the remainder. That is option (A).
$z=-i$ makes $z+i$ zero, so $2i$ is the remainder when you divide by $z+i$. That is option (C), but the question divides by $z-i$.
The remainder theorem: when $P(z)$ is divided by $z-a$, the remainder is $P(a)$. Here the divisor is $z-i$, so $a=i$.
You need the powers of $i$.
$$i^2=-1,\qquad i^3=i^2\times i=-i$$
Substitute term by term.
$$P(i)=i^3-i\times i^2-i-i$$
$$P(i)=(-i)-i(-1)-i-i$$
$$=-i+i-i-i=-2i$$
Check by division. The quotient is $z^2-1$:
$$(z-i)(z^2-1)-2i$$
$$=z^3-iz^2-z+i-2i$$
$$=z^3-iz^2-z-i\ \checkmark$$
By the remainder theorem, the remainder on division by $z-i$ is $P(i)$.
$$P(i)=i^3-i\times i^2-i-i$$
$$=-i+i-i-i=-2i$$
How Queensland students answered in 2021
From the QCAA Specialist Mathematics subject report for 2021. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it