Question 8
The imaginary part of $\left(\operatorname{cis}\left(\frac{\pi}{8}\right)\right)^{-2}$ is
Not quite. $-6.83$ is $\left(i\sin\frac{\pi}{8}\right)^{-2}=-\frac{1}{\sin^2\frac{\pi}{8}}$. That raises the $i\sin$ term to the power on its own. A power of a sum is not the sum of the powers. De Moivre’s theorem puts the power on the angle instead: $\operatorname{cis}\left(-\frac{\pi}{4}\right)$. In 2021, 15.8% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2021, 64.5% of Queensland students chose (B).
Not quite. $0.71$ is $\sin\frac{\pi}{4}$, the imaginary part of $\left(\operatorname{cis}\frac{\pi}{8}\right)^{2}$. The power is $-2$, so the angle is $-\frac{\pi}{4}$ and the imaginary part is $\sin\left(-\frac{\pi}{4}\right)\approx-0.71$. In 2021, 13.6% of Queensland students chose (C).
Not quite. $1.17$ is $\left(\cos\frac{\pi}{8}\right)^{-2}=\frac{1}{\cos^2\frac{\pi}{8}}$. That raises the real part to the power on its own, and the imaginary part is never found. De Moivre’s theorem gives $\operatorname{cis}\left(-\frac{\pi}{4}\right)$, whose imaginary part is $\sin\left(-\frac{\pi}{4}\right)\approx-0.71$. In 2021, 5.4% of Queensland students chose (D).
The grey point is $z=\operatorname{cis}\left(\frac{\pi}{8}\right)$. The orange point is $z^n$, and the dashed blue line is its imaginary part.
This is the power in the question. The point has turned clockwise to $-\frac{\pi}{4}$, below the real axis, so the imaginary part is negative: about $-0.71$.
A power of $+2$ turns the point anticlockwise to $\frac{\pi}{4}$, which gives $0.71$, option (C). The question has a power of $-2$.
Each step in the power turns the point another $\frac{\pi}{8}$ round the circle. Positive powers turn anticlockwise and negative powers turn clockwise. The modulus stays at $1$.
De Moivre’s theorem says a power of $\operatorname{cis}\theta$ multiplies the angle. It works for negative integer powers too.
$$\left(\operatorname{cis}\theta\right)^n=\operatorname{cis}(n\theta)$$
Here $\theta=\frac{\pi}{8}$ and $n=-2$.
$$\left(\operatorname{cis}\tfrac{\pi}{8}\right)^{-2}=\operatorname{cis}\left(-\tfrac{\pi}{4}\right)$$
$$=\cos\left(-\tfrac{\pi}{4}\right)+i\sin\left(-\tfrac{\pi}{4}\right)$$
The imaginary part is the real number multiplying $i$.
$$\sin\left(-\tfrac{\pi}{4}\right)=-\tfrac{\sqrt2}{2}\approx-0.71$$
To check on your calculator, enter $\left(e^{i\pi/8}\right)^{-2}$ in complex mode with angles in radians. It returns $0.707-0.707i$.
By De Moivre’s theorem, the power $-2$ multiplies the angle.
$$\begin{aligned}\left(\operatorname{cis}\tfrac{\pi}{8}\right)^{-2}&=\operatorname{cis}\left(-\tfrac{\pi}{4}\right)\\&=\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}i\end{aligned}$$
The imaginary part is $-\frac{\sqrt2}{2}\approx-0.71$.
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.
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