# Thread: HSC 1991 I think! Can anyone help?

2. ## Re: HSC 1991 I think! Can anyone help?

May i know where u got this question from?

3. ## Re: HSC 1991 I think! Can anyone help?

i) $z^7 - 1 = 0$

Since $z \neq 1$,

$\frac{z^7 - 1}{z-1} = 0$

You can use long division and then divide by $z^3$ (since $z \neq 0$) to obtain the desired result. It's rather unintuitive, though.
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ii) Consider the expansions of

$\left( z + \frac{1}{z}\right)^2 \,and\, \left( z + \frac{1}{z}\right)^3$.
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iii) There are six solutions to the equation in part i). By performing the substitution in ii), you've reduced it down to three, by making three of those solutions the same as the other three.

Find these three unique solutions and use the product of roots and the properties of $\cos x$ in the first and second quadrants to obtain the answer.

$e.g.\, \cos \frac{4\pi}{7} = - \cos \frac{3\pi}{7}$

4. ## Re: HSC 1991 I think! Can anyone help?

Originally Posted by HeroWise
May i know where u got this question from?
HSC 1992 Q7

If you are interested in these type of questions, I'd recommend checking out coroneos (The supplementary book), it has a lot of similar/Interesting questions.

5. ## Re: HSC 1991 I think! Can anyone help?

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