Question 19
The three cube roots of a complex number $w$ are plotted on an Argand plane and then joined by straight lines to form a triangle with an area of $3\sqrt{3}$ units2.
One of the cube roots of $w$ has equal real and imaginary parts.
Determine all possible values of $w$. Leave your answers in Cartesian form.
[6 marks]What the question doesn’t tell you
Have a go at the question cold first. Then open this to see the steps the question expects you to find for yourself.
- The three cube roots of any non-zero complex number have the same modulus and sit $\tfrac{2\pi}{3}$ apart. So they are the corners of an equilateral triangle centred at the origin. The question never says “equilateral” or “circle”.
- To use the area, you have to invent the unknown yourself: the common modulus $r$ of the roots. You are not given a side length or a vertex.
- The modulus that comes out of the area belongs to the roots, not to $w$. You need $\lvert w\rvert=r^3$.
- “Equal real and imaginary parts” puts that root on the line $\operatorname{Im}(z)=\operatorname{Re}(z)$. The two parts can both be negative, so the argument is $\tfrac{\pi}{4}$ or $-\tfrac{3\pi}{4}$, and there are two values of $w$. The question says “all possible values” but not how many there are.
- You never need to find the roots of $w$ from $w$. One root is enough: cube it. This runs the usual “find the cube roots of $w$” task backwards.
- There is a shortcut for the second value: it is the negative of the first. If $z$ is on the line, so is $-z$, and $(-z)^3=-z^3$.
Three cube roots on a circle, and the one on the line cubes back to $w$
Watch the roots settle on a circle and split the triangle into three equal pieces. Then watch the root on the line cube out to $w$, and a second triangle give a second $w$.
Both panels are drawn to scale, with equal scales on the axes. Cubing a root takes its modulus from 2 to 8 and triples its argument, and the second triangle is the first turned through $\pi$ about $O$.
Choose $w$ and watch its cube roots
Set the modulus of the roots and the argument of $w$. The triangle of cube roots follows. You need an area of $3\sqrt3$ and one root on the dashed line. Two values of $w$ do both. Find them, then switch routes to watch each method reach the same $w$.
Both conditions hold. The area is $3\sqrt3$, and the root $2\operatorname{cis}\left(\tfrac{\pi}{4}\right)=\sqrt2+\sqrt2\,i$ is on the line. Cubing it gives $w=8\operatorname{cis}\left(\tfrac{3\pi}{4}\right)=-4\sqrt2+4\sqrt2\,i$.
This works too. The root on the line is $-\sqrt2-\sqrt2\,i=2\operatorname{cis}\left(-\tfrac{3\pi}{4}\right)$, with both parts negative. Cubing it gives $w=8\operatorname{cis}\left(-\tfrac{\pi}{4}\right)=4\sqrt2-4\sqrt2\,i$, the negative of the other value.
A root is on the dashed line, so its real and imaginary parts are equal. The area is not $3\sqrt3$ yet, so change the modulus.
The area is exactly $3\sqrt3$, so each root has modulus 2 and $\lvert w\rvert=2^3=8$. Now turn $w$ until one root lands on the dashed line.
The area is more than $3\sqrt3$, so the roots are too far from $O$. Make the modulus smaller.
The area is less than $3\sqrt3$, so the roots are too close to $O$. Make the modulus larger.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Start from what the three cube roots of any complex number look like. Write $w=R\operatorname{cis}(\varphi)$ with $R>0$. You want the solutions of $z^3=R\operatorname{cis}(\varphi)$.
Every piece of information is about the roots, not about $w$. Also $w\neq0$, because the only cube root of 0 is 0 and that gives no triangle.
Use the method for the $n$th roots of any complex number. Here $n=3$, and the right side is already in polar form: $R\operatorname{cis}(\varphi)$. Use De Moivre’s theorem to find the first root.
$$z_0=R^{\frac13}\operatorname{cis}\left(\frac{\varphi}{3}\right)$$
There are 3 solutions, spaced $\tfrac{2\pi}{3}$ (or $\tfrac{360^\circ}{3}=120^\circ$) apart. Add $\tfrac{2\pi}{3}$ to the argument until you are back at the first root. Every root has the same modulus, $r=R^{\frac13}$.
$$\begin{aligned}z_0&=r\operatorname{cis}\left(\frac{\varphi}{3}\right)\\z_1&=r\operatorname{cis}\left(\frac{\varphi}{3}+\frac{2\pi}{3}\right)\\z_2&=r\operatorname{cis}\left(\frac{\varphi}{3}+\frac{4\pi}{3}\right)\end{aligned}$$
What this tells you about the triangle: the roots all have modulus $r$ and are spaced $\tfrac{2\pi}{3}$ apart, so the angle at $O$ between any two neighbouring $r$’s is $\tfrac{2\pi}{3}$.
Write the area in terms of the modulus $r$. Join $O$ to each vertex, then use $\tfrac12ab\sin(C)$ on one piece.
$$\begin{aligned}\text{one piece}&=\tfrac12\times r\times r\times\sin\left(\tfrac{2\pi}{3}\right)\\\sin\left(\tfrac{2\pi}{3}\right)&=\sin\left(\pi-\tfrac{\pi}{3}\right)\\&=\sin\left(\tfrac{\pi}{3}\right)=\tfrac{\sqrt3}{2}\\\text{one piece}&=\tfrac12r^2\times\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{4}r^2\\\text{Area}&=3\times\tfrac{\sqrt3}{4}r^2\\&=\frac{3\sqrt3}{4}r^2\end{aligned}$$
The area is the only number you are given, so write it in terms of the one thing that sets the triangle’s size: the modulus $r$. Joining $O$ to each vertex cuts the triangle into three pieces, each with two sides $r$ and the angle $\tfrac{2\pi}{3}$ between them. The three angles at $O$ add to $2\pi$, so the pieces fill the triangle exactly.
Marker: correctly determines the area of the triangle formed by the cube roots in terms of their modulus
Set the area equal to $3\sqrt3$ and solve for $r$.
$$\begin{aligned}\frac{3\sqrt3}{4}r^2&=3\sqrt3\\r^2&=3\sqrt3\times\frac{4}{3\sqrt3}\\&=4\\r&=2\quad(r>0)\end{aligned}$$
There is now one equation in one unknown, so multiply both sides by $\tfrac{4}{3\sqrt3}$. Reject $r=-2$, because a modulus is positive.
This is the modulus of each cube root, not of $w$. The modulus of $w$ itself is $\lvert w\rvert=r^3=2^3=8$, and you need that when you cube a root.
Marker: correctly determines the modulus of the cube roots
Call this root $z_{\text{eq}}=a+ai$ with $a\in R$. Use its modulus from Step 2 to find $a$.
$$\begin{aligned}\lvert z_{\text{eq}}\rvert&=\sqrt{a^2+a^2}\\&=\sqrt{2a^2}\\&=\sqrt2\,\sqrt{a^2}\\&=\sqrt2\,\lvert a\rvert\quad\left(\text{since }\sqrt{a^2}=\lvert a\rvert\right)\\\sqrt2\,\lvert a\rvert&=2\\\lvert a\rvert&=\frac{2}{\sqrt2}=\sqrt2\\a&=\sqrt2\ \text{ or }\ a=-\sqrt2\end{aligned}$$
$$\sqrt{a^2}=\lvert a\rvert$$
“Equal real and imaginary parts” fixes the direction of this root up to sign, and Step 2 fixes its distance from $O$. Together they pin it down. Here $a\neq0$ because $\lvert z_{\text{eq}}\rvert=2$. The condition says the parts are equal, not that they are positive, so both signs satisfy it and both stay.
Find the argument for each sign. With $a=\sqrt2$, $z_{\text{eq}}=\sqrt2+\sqrt2\,i$ is in the first quadrant. With $a=-\sqrt2$, $z_{\text{eq}}=-\sqrt2-\sqrt2\,i$ is in the third quadrant, where $-\pi<\theta<-\tfrac{\pi}{2}$.
$$\begin{aligned}\text{first quadrant: }\tan(\theta)&=\tfrac{\sqrt2}{\sqrt2}=1\\\theta&=\tfrac{\pi}{4}\\\text{third quadrant: }\tan(\theta)&=1\\\theta&=-\pi+\tfrac{\pi}{4}=-\tfrac{3\pi}{4}\end{aligned}$$
Write both roots in polar form.
$$\begin{aligned}&z_{\text{eq}}=2\operatorname{cis}\left(\tfrac{\pi}{4}\right)\\&\quad\text{or}\\&z_{\text{eq}}=2\operatorname{cis}\left(-\tfrac{3\pi}{4}\right)\end{aligned}$$
Marker: recognises that the cube root may have argument $\tfrac{\pi}{4}$ or $-\tfrac{3\pi}{4}$
Cube the first-quadrant root. De Moivre’s theorem cubes the modulus and triples the argument.
$$\begin{aligned}w&=z_{\text{eq}}^3\\&=\left(2\operatorname{cis}\left(\tfrac{\pi}{4}\right)\right)^3\\&=2^3\operatorname{cis}\left(3\times\tfrac{\pi}{4}\right)\\&=8\operatorname{cis}\left(\tfrac{3\pi}{4}\right)\end{aligned}$$
$z_{\text{eq}}$ is a cube root of $w$, so by definition $z_{\text{eq}}^3=w$. Cubing one known root gives $w$ directly. You never need the other two roots.
The question asks for Cartesian form, so expand the $\operatorname{cis}$.
$$\begin{aligned}w&=8\left(\cos\left(\tfrac{3\pi}{4}\right)+i\sin\left(\tfrac{3\pi}{4}\right)\right)\\\cos\left(\tfrac{3\pi}{4}\right)&=-\cos\left(\tfrac{\pi}{4}\right)=-\tfrac{\sqrt2}{2}\\\sin\left(\tfrac{3\pi}{4}\right)&=\sin\left(\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}\\w&=8\left(-\tfrac{\sqrt2}{2}\right)+8\left(\tfrac{\sqrt2}{2}\right)i\\&=-4\sqrt2+4\sqrt2\,i\end{aligned}$$
The angle $\tfrac{3\pi}{4}=\pi-\tfrac{\pi}{4}$ is in the second quadrant with reference angle $\tfrac{\pi}{4}$. There cosine is negative and sine is positive.
The polar line $w=8\operatorname{cis}\left(\tfrac{3\pi}{4}\right)$ already earns this mark, because polar form is accepted for one value.
Marker: correctly determines $w$ in polar or Cartesian form for one cube root
The other branch from Step 3 gives a second, different $w$. Treat it the same way.
$$\begin{aligned}w&=\left(2\operatorname{cis}\left(-\tfrac{3\pi}{4}\right)\right)^3\\&=2^3\operatorname{cis}\left(3\times\left(-\tfrac{3\pi}{4}\right)\right)\\&=8\operatorname{cis}\left(-\tfrac{9\pi}{4}\right)\end{aligned}$$
$-\tfrac{9\pi}{4}$ is outside $(-\pi,\pi]$, so add $2\pi$. That does not move the point.
$$\begin{aligned}-\tfrac{9\pi}{4}+2\pi&=-\tfrac{\pi}{4}\\\Rightarrow w&=8\operatorname{cis}\left(-\tfrac{\pi}{4}\right)\end{aligned}$$
Expand the $\operatorname{cis}$ to get Cartesian form.
$$\begin{aligned}w&=8\left(\cos\left(-\tfrac{\pi}{4}\right)+i\sin\left(-\tfrac{\pi}{4}\right)\right)\\\cos\left(-\tfrac{\pi}{4}\right)&=\cos\left(\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}\\\sin\left(-\tfrac{\pi}{4}\right)&=-\sin\left(\tfrac{\pi}{4}\right)=-\tfrac{\sqrt2}{2}\\w&=8\left(\tfrac{\sqrt2}{2}\right)+8\left(-\tfrac{\sqrt2}{2}\right)i\\&=4\sqrt2-4\sqrt2\,i\end{aligned}$$
$-\tfrac{\pi}{4}$ is in the fourth quadrant, where cosine is positive and sine is negative. Cosine is even and sine is odd.
A response giving one value of $w$ only does not gain this mark.
Marker: correctly determines both values of $w$ in Cartesian form
Each value works by construction: its cube roots include $z_{\text{eq}}$, so all three have modulus 2 and the triangle has area $3\sqrt3$. Nothing else works. Any root with equal parts must have modulus 2 (Step 2) and argument $\tfrac{\pi}{4}$ or $-\tfrac{3\pi}{4}$ (Step 3). Cubing those two roots gives only these two values of $w$.
“All possible values” means two things: each value works, and nothing else does.
Set the working out in order, from the area to the modulus to both roots on the line to both values of $w$.
Setting the working out in order like this, with the whole question attempted, is the last mark.
Marker: shows logical organisation, having fully attempted the question
As in Method 1, describe the roots first. Write $w=R\operatorname{cis}(\varphi)$ with $R>0$. With $n=3$, De Moivre’s theorem gives the first root, and the other two come from adding $\tfrac{2\pi}{3}$ to the argument until you are back at $z_0$. Every root has modulus $r$.
$$\begin{aligned}z_0&=R^{\frac13}\operatorname{cis}\left(\tfrac{\varphi}{3}\right)\\r&=R^{\frac13}\end{aligned}$$
The information is about the roots, so describe them first. Here $w\neq0$, or there is no triangle. All three roots lie on the circle of radius $r$ centred at $O$, and adjacent roots are $\tfrac{2\pi}{3}$ apart. They form an equilateral triangle centred at $O$.
As in Method 1, Step 1, join $O$ to each vertex. That splits the triangle into three pieces, each with two sides $r$ and included angle $\tfrac{2\pi}{3}$.
$$\begin{aligned}\text{Area}&=3\times\tfrac12r^2\sin\left(\tfrac{2\pi}{3}\right)\\&=3\times\tfrac12r^2\times\tfrac{\sqrt3}{2}\\&=\frac{3\sqrt3}{4}r^2\end{aligned}$$
Marker: correctly determines the area of the triangle formed by the cube roots in terms of their modulus
Set the area expression equal to the given area and solve for $r$.
$$\begin{aligned}\frac{3\sqrt3}{4}r^2&=3\sqrt3\\r^2&=3\sqrt3\times\frac{4}{3\sqrt3}=4\\r&=2\quad(r>0)\end{aligned}$$
That is one equation in the one unknown $r$, so multiply both sides by $\tfrac{4}{3\sqrt3}$.
This is the modulus of each cube root. The modulus of $w$ is $r^3=8$.
Marker: correctly determines the modulus of the cube roots
Let $z_1=a+ai$ with $a\in R$. Use the modulus from Step 2 to find $a$.
$$\begin{aligned}\lvert z_1\rvert&=\sqrt{a^2+a^2}\\&=\sqrt2\,\sqrt{a^2}\\&=\sqrt2\,\lvert a\rvert=2\\\lvert a\rvert&=\frac{2}{\sqrt2}=\sqrt2\\a&=\pm\sqrt2\end{aligned}$$
$$\begin{aligned}&z_1=\sqrt2+\sqrt2\,i\\&\quad\text{or}\\&z_1=-\sqrt2-\sqrt2\,i\end{aligned}$$
This method stays Cartesian, so keep the root as $a+ai$ instead of converting to polar. The parts can both be negative, so keep both signs.
These are the points with arguments $\tfrac{\pi}{4}$ (first quadrant) and $-\tfrac{3\pi}{4}$ (third quadrant).
Writing down both roots on the line is this method’s version of recognising both arguments.
Marker: recognises that the cube root may have argument $\tfrac{\pi}{4}$ or $-\tfrac{3\pi}{4}$
Cube the root, $w=z_1^3$. Factor out $\sqrt2$ first.
$$\begin{aligned}w&=\left(\sqrt2+\sqrt2\,i\right)^3\\&=\left(\sqrt2(1+i)\right)^3\\&=\left(\sqrt2\right)^3(1+i)^3\\\left(\sqrt2\right)^3&=\sqrt2\times\sqrt2\times\sqrt2=2\sqrt2\end{aligned}$$
The root is a binomial, so expand it directly. Factoring out $\sqrt2$ first keeps the numbers small.
Expand $(1+i)^3$ with the binomial coefficients 1, 3, 3, 1. Use $i^2=-1$ and $i^3=i^2\times i=-i$.
$$\begin{aligned}(1+i)^3&=1+3i+3i^2+i^3\\&=1+3i-3-i\\&=-2+2i\end{aligned}$$
Multiply by $2\sqrt2$.
$$\begin{aligned}w&=2\sqrt2\,(-2+2i)\\&=-4\sqrt2+4\sqrt2\,i\end{aligned}$$
Marker: correctly determines $w$ in polar or Cartesian form for one cube root
Write the third-quadrant root as the negative of the first, then cube it.
$$\begin{aligned}w&=\left(-\sqrt2-\sqrt2\,i\right)^3\\&=\left(-\left(\sqrt2+\sqrt2\,i\right)\right)^3\\&=(-1)^3\left(\sqrt2+\sqrt2\,i\right)^3\\&=-\left(-4\sqrt2+4\sqrt2\,i\right)\\&=4\sqrt2-4\sqrt2\,i\end{aligned}$$
$-\sqrt2-\sqrt2\,i=-\left(\sqrt2+\sqrt2\,i\right)$. An odd power keeps the minus sign, so its cube is the negative of the first $w$. You do not need a second expansion.
A response giving one value of $w$ only does not gain this mark.
Marker: correctly determines both values of $w$ in Cartesian form
There are no other values. A root with equal parts must be $\pm\left(\sqrt2+\sqrt2\,i\right)$, and Step 3 found both. Set the working out in order, from the area to both values of $w$.
Check the modulus. Both values should have $\lvert w\rvert=r^3=8$.
$$\begin{aligned}\lvert w\rvert&=\sqrt{\left(4\sqrt2\right)^2+\left(4\sqrt2\right)^2}\\&=\sqrt{32+32}=\sqrt{64}\\&=8=2^3\ \checkmark\end{aligned}$$
A modulus check catches most expansion slips. The two values differ only in the signs of their parts, so one calculation covers both.
Setting the working out in order, with the whole question attempted, is the last mark.
Marker: shows logical organisation, having fully attempted the question
Each phrase of the question gave you something
The engine here is inversion. The usual task is to take $w$, find its cube roots and plot them. This question runs that backwards. The data arrive as a triangle area and a Cartesian condition on one root. Nothing in the stem tells you what to do with them. You have to reach for assumed knowledge on your own (the roots form an equilateral triangle, and area is $\tfrac12ab\sin(C)$), switch from “equal parts” to polar arguments, and spot a hidden second branch.
If I had to guess where you might lose marks…
- First quadrant only. You read “equal real and imaginary parts” as positive parts, take $\tfrac{\pi}{4}$ alone and give only $w=-4\sqrt2+4\sqrt2\,i$. That loses the third and fifth marks.
- One piece instead of three. You set $\tfrac{\sqrt3}{4}r^2=3\sqrt3$, by forgetting the factor of 3 or by treating $r$ as the side length. Then $r=2\sqrt3$, so $\lvert w\rvert=24\sqrt3$ and $w=-12\sqrt6+12\sqrt6\,i$ or $12\sqrt6-12\sqrt6\,i$, with parts of about $\pm29.3939$.
- Modulus not cubed. You treat 2 as $\lvert w\rvert$ and get $w=2\operatorname{cis}\left(\tfrac{3\pi}{4}\right)=-\sqrt2+\sqrt2\,i$ or $\sqrt2-\sqrt2\,i$.
- Argument not tripled. You cube the modulus but keep the argument, and get $w=8\operatorname{cis}\left(\tfrac{\pi}{4}\right)=4\sqrt2+4\sqrt2\,i$ or $8\operatorname{cis}\left(-\tfrac{3\pi}{4}\right)=-4\sqrt2-4\sqrt2\,i$.
- An $i^3$ slip on the binomial route. You take $i^3=i$ and get $(1+i)^3=-2+4i$, so $w=-4\sqrt2+8\sqrt2\,i$. Its modulus is $4\sqrt{10}$, not 8, so a modulus check catches it.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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