Question 6
Subsets of the complex plane like this one are now taught in Unit 2, so this won’t be directly examined in your Units 3 and 4 external exam. It is still good practice with modulus and the Argand diagram.
The subset of the complex plane that represents $|z|=|z-2|$ for $z\in\mathbb{C}$ is
Not quite. A horizontal line through $i$ is the set of points the same distance from $0$ and $2i$, which is $|z|=|z-2i|$. In $|z-2|$ the $2$ is real, so the second point sits on the real axis and the line is vertical. In 2021, 17.5% of Queensland students chose (A).
Not quite. A horizontal line through $-i$ is the set of points the same distance from $0$ and $-2i$, which is $|z|=|z+2i|$. In $|z-2|$ the second point is $2$, on the real axis. In 2021, 8.8% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2021, 57.5% of Queensland students chose (C).
Not quite. The line $\operatorname{Re}(z)=-1$ is halfway between $0$ and $-2$, so it shows $|z|=|z+2|$. In the question, $|z-2|=|z-(+2)|$ is the distance from $+2$, which puts the line halfway between $0$ and $2$ instead. In 2021, 15.6% of Queensland students chose (D).
Drag the orange point. The black segment is its distance from $0$. The blue segment is its distance from the second point, which you choose with the buttons. The dashed line is every point the same distance from both.
The two distances are equal, so this point is on the dashed line .
This point is closer to than to . Drag it onto the dashed line to make the distances equal.
This is the question’s condition. The points the same distance from $0$ and $2$ form the vertical line $\operatorname{Re}(z)=1$, which is option (C).
This is $|z|=|z+2|$, the distance from $-2$. Its line is $\operatorname{Re}(z)=-1$, option (D). The question measures from $+2$.
This is $|z|=|z-2i|$. The second point $2i$ is on the imaginary axis, so the line is horizontal, $\operatorname{Im}(z)=1$, option (A). In the question the $2$ is real.
This is $|z|=|z+2i|$. The second point $-2i$ is on the imaginary axis, so the line is horizontal, $\operatorname{Im}(z)=-1$, option (B). In the question the $2$ is real.
Read each modulus as a distance. $|z|=|z-0|$ is the distance from $z$ to $0$, and $|z-2|$ is the distance from $z$ to the point $2$ on the real axis.
So the question wants every point the same distance from $0$ and $2$. Those points form the perpendicular bisector of the segment from $0$ to $2$. It passes through the midpoint $1$ and is vertical.
Check with algebra. Let $z=x+yi$.
$$x^2+y^2=(x-2)^2+y^2$$
$$0=-4x+4\ \Rightarrow\ x=1$$
This is the vertical line $\operatorname{Re}(z)=1$, one unit right of the imaginary axis.
$|z|$ is the distance from $z$ to $0$ and $|z-2|$ is the distance from $z$ to $2$. With $z=x+yi$,
$$x^2+y^2=(x-2)^2+y^2\ \Rightarrow\ x=1$$
The subset is the vertical line $\operatorname{Re}(z)=1$, the perpendicular bisector of $0$ and $2$.
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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