Question 19

You start by writing down what "fifth root of unity" means. It is a solution of $w^{5}=1$, and you are told it is not the real one.

$$w^{5}=1,\quad w\notin\mathbb{R}\quad\ldots(1)$$

This mark can be implied if you use $w^{5}=1$ later in your working, but it costs you nothing to write it down first.

Rearrange to $w^{5}-1=0$. You know $w=1$ is the real solution, so by the factor theorem $w-1$ is a factor.

$$w^{5}-1=(w-1)\,Q(w)$$

You also get this mark if you set up the division $\frac{w^{5}-1}{w-1}$, or if you say the roots you want come from a polynomial of that form. It can be implied by your later working, and follow-through marks are allowed.

Now you find $Q(w)$. It is a quartic, so write it with unknown coefficients and equate them.

$$(w-1)\left(w^{4}+aw^{3}+bw^{2}+cw+1\right)=w^{5}-1$$

$$\begin{aligned}w&:\ 1-c=0\Rightarrow c=1\\ w^{2}&:\ c-b=0\Rightarrow b=1\\ w^{3}&:\ b-a=0\Rightarrow a=1\end{aligned}$$

$$w^{5}-1=(w-1)\left(w^{4}+w^{3}+w^{2}+w+1\right)$$

Long division gets you the same quartic if you prefer it, and the QCAA's second method does exactly that. You do not have to use long division for this mark.

This is the step you need to see coming. Because $w\neq1$, the factor $w-1$ is not zero, so the quartic must be zero for all four non-real roots.

$$(w-1)\left(w^{4}+w^{3}+w^{2}+w+1\right)=0$$

$$w^{4}+w^{3}+w^{2}+w+1=0\quad\ldots(2)$$

Writing it as $w^{4}+w^{3}+w^{2}+w=-1$ is just as good. That version is the one you will actually use in step 6.

Now you expand the expression you were given. The powers go past 5, so use (1) to bring them back down.

$$\begin{aligned}w^{3}(1+w)(1+w^{3})&=\left(w^{3}+w^{4}\right)\left(1+w^{3}\right)\\&=w^{3}+w^{6}+w^{4}+w^{7}\end{aligned}$$

$$w^{6}=w^{5}\cdot w=w\qquad w^{7}=w^{5}\cdot w^{2}=w^{2}$$

$$w^{3}(1+w)(1+w^{3})=w^{4}+w^{3}+w^{2}+w$$

You could also add and subtract $w^{5}$ terms instead, which the QCAA's second method does. Either way you are using $w^{5}=1$ to tidy the powers.

You start by writing down what you are told.

$$w^{5}=1,\quad w\notin\mathbb{R}\quad\ldots(1)$$

This mark can be implied by your later working.

$w=1$ is the real solution of $w^{5}-1=0$, so $w-1$ is a factor. The four roots you want are the roots of what is left over.

$$Q(w)=\frac{w^{5}-1}{w-1}$$

Setting up this division is enough for the mark. It can be implied by your later working, and follow-through marks are allowed.

Now you carry out the long division. Each line below is one step of it, where you take out another multiple of $w-1$.

$$\begin{aligned}w^{5}-1&=w^{4}(w-1)+\left(w^{4}-1\right)\\w^{4}-1&=w^{3}(w-1)+\left(w^{3}-1\right)\\w^{3}-1&=w^{2}(w-1)+\left(w^{2}-1\right)\\w^{2}-1&=w(w-1)+(w-1)\\w-1&=1\cdot(w-1)+0\end{aligned}$$

Adding up the multiples gives you the quartic.

$$w^{5}-1=(w-1)\left(w^{4}+w^{3}+w^{2}+w+1\right)$$

You do not have to use long division for this mark. Equating coefficients, as in method 1, is just as good.

Because $w\neq1$, the factor $w-1$ is not zero, so the quartic must be zero for all four non-real roots.

$$w^{4}+w^{3}+w^{2}+w+1=0\quad\ldots(2)$$

Expand the expression. This time, instead of reducing $w^{6}$ and $w^{7}$, you add and subtract $w^{5}$ so that the quartic appears.

$$\begin{aligned}w^{3}(1+w)(1+w^{3})&=w^{7}+w^{6}+w^{4}+w^{3}\\&=w^{7}+w^{6}+w^{5}+w^{4}+w^{3}-w^{5}\\&=w^{3}\left(w^{4}+w^{3}+w^{2}+w+1\right)-w^{5}\end{aligned}$$

This mark can be implied by your later working. Writing it as $w^{2}+w+w^{5}+w^{4}+w^{3}-w^{5}$ is also accepted.

You expand first and use $w^{5}=1$ to bring the powers down.

$$\begin{aligned}w^{3}(1+w)(1+w^{3})&=\left(w^{3}+w^{4}\right)\left(1+w^{3}\right)\\&=w^{3}+w^{6}+w^{4}+w^{7}\\&=w^{3}+w+w^{4}+w^{2}\end{aligned}$$

Put the terms in order and look at them again. Each term is the one before multiplied by $w$, so you have a geometric sequence.

$$w+w^{2}+w^{3}+w^{4}$$

Stating the sum formula $S_{n}=\frac{t_{1}\left(r^{n}-1\right)}{r-1}$ also earns this mark. It can be implied by your later working, and follow-through marks are allowed.

Write down the three values you need for the formula.

$$t_{1}=w,\qquad r=w,\qquad n=4$$

This mark can be implied by your later working.

Substitute into the sum formula and expand the numerator. You are allowed to use the formula because $r=w\neq1$, which is exactly what $w\notin\mathbb{R}$ tells you.

$$S_{4}=\frac{w\left(w^{4}-1\right)}{w-1}=\frac{w^{5}-w}{w-1}$$

Writing it as $\frac{w-w^{5}}{1-w}$ is also accepted.

Now you use the fact that $w$ is a fifth root of unity, so $w^{5}=1$.

$$w^{3}(1+w)(1+w^{3})=\frac{1-w}{w-1}$$

You start the same way. Write down that $w$ solves $w^{5}=1$ and that it is not the real root.

$$w^{5}=1,\quad w\notin\mathbb{R}$$

This mark can be implied by your later working.

This time you find the roots themselves. The fifth roots of unity sit on the unit circle, spaced $\frac{2\pi}{5}$ apart. You throw out $w=1$, which leaves you with four.

$$w=\operatorname{cis}\left(\pm\frac{2\pi}{5}\right)\ \text{and}\ \operatorname{cis}\left(\pm\frac{4\pi}{5}\right)$$

An Argand diagram with a modulus of 1 and these four arguments marked also earns the mark. You still get it if you leave the real root in, and follow-through marks are allowed.

Expand the expression and use $w^{5}=1$ to bring the powers down, exactly as in method 1.

$$\begin{aligned}w^{3}(1+w)(1+w^{3})&=w^{3}+w^{4}+w^{6}+w^{7}\\&=w+w^{2}+w^{3}+w^{4}\quad\ldots(1)\end{aligned}$$

Now you substitute a root into (1). Take $w=\operatorname{cis}\left(\frac{2\pi}{5}\right)$, so each power just multiplies the argument.

$$\begin{aligned}w+w^{2}+w^{3}+w^{4}&=\operatorname{cis}\tfrac{2\pi}{5}+\operatorname{cis}\tfrac{4\pi}{5}+\operatorname{cis}\tfrac{6\pi}{5}+\operatorname{cis}\tfrac{8\pi}{5}\\&=\cos\tfrac{2\pi}{5}+\cos\tfrac{4\pi}{5}+\cos\tfrac{6\pi}{5}+\cos\tfrac{8\pi}{5}\\&\quad+i\left(\sin\tfrac{2\pi}{5}+\sin\tfrac{4\pi}{5}+\sin\tfrac{6\pi}{5}+\sin\tfrac{8\pi}{5}\right)\\&=2\cos\tfrac{2\pi}{5}-2\cos\tfrac{\pi}{5}\end{aligned}$$

The sine terms cancel because $\frac{6\pi}{5}$ and $\frac{8\pi}{5}$ are the reflections of $\frac{4\pi}{5}$ and $\frac{2\pi}{5}$ below the real axis. The cosines pair up instead, and you use $\cos\frac{4\pi}{5}=-\cos\frac{\pi}{5}$. Any of the other three roots gives you the same result.

You now need exact values for $\cos\frac{\pi}{5}$ and $\cos\frac{2\pi}{5}$, and you have to build them yourself. Let $\theta=\frac{\pi}{10}$, so that $5\theta=\frac{\pi}{2}$ and $2\theta=\frac{\pi}{2}-3\theta$.

$$\begin{aligned}\sin(2\theta)&=\sin\left(\tfrac{\pi}{2}-3\theta\right)=\cos(3\theta)\\2\sin(\theta)\cos(\theta)&=4\cos^{3}(\theta)-3\cos(\theta)\\0&=\cos(\theta)\left(4\cos^{2}(\theta)-2\sin(\theta)-3\right)\end{aligned}$$

Since $\cos(\theta)\neq0$, you can divide it out and swap $\cos^{2}(\theta)$ for $1-\sin^{2}(\theta)$.

$$\begin{aligned}4\sin^{2}(\theta)+2\sin(\theta)-1&=0\\\sin(\theta)&=\frac{-1\pm\sqrt{5}}{4}\\\sin\left(\tfrac{\pi}{10}\right)&=\frac{-1+\sqrt{5}}{4}\quad\text{as }\theta\text{ is acute}\end{aligned}$$

Then the double angle identities give you both cosines.

$$\begin{aligned}\cos\left(\tfrac{\pi}{5}\right)&=1-2\sin^{2}\left(\tfrac{\pi}{10}\right)=\frac{1+\sqrt{5}}{4}\\\cos\left(\tfrac{2\pi}{5}\right)&=2\cos^{2}\left(\tfrac{\pi}{5}\right)-1=\frac{\sqrt{5}-1}{4}\end{aligned}$$

You start by writing down what you are told.

$$w^{5}=1,\quad w\notin\mathbb{R}$$

This mark can be implied by your later working.

Expand the expression you were given.

$$\begin{aligned}w^{3}(1+w)(1+w^{3})&=\left(w^{3}+w^{4}\right)\left(1+w^{3}\right)\\&=w^{7}+w^{6}+w^{4}+w^{3}\end{aligned}$$

Use $w^{5}=1$ to bring the powers down.

$$w^{7}+w^{6}+w^{4}+w^{3}=w^{4}+w^{3}+w^{2}+w$$

Follow-through marks are allowed here.

The roots of unity always add to zero, because they sit evenly around the unit circle. This is what the animation shows. The five fifth roots are $w$, $w^{2}$, $w^{3}$, $w^{4}$ and $w^{5}$, so you can write their sum like this.

$$w^{5}+w^{4}+w^{3}+w^{2}+w=0$$

Replace $w^{5}$ with 1 and move it across.

$$\begin{aligned}w^{4}+w^{3}+w^{2}+w+1&=0\\w^{4}+w^{3}+w^{2}+w&=-1\end{aligned}$$

Follow-through marks are allowed here.