 July 12, 2020, 11:21:04 pm

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#### RuiAce

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« Reply #2475 on: May 24, 2020, 08:09:07 pm »
+2
Thanks a lot for your help!!! I get it now. Also stuck with part (c) of another question. For these types of questions, I know how to solve the LHS but for the RHS I'm not sure. I tried changing the roots from the equation in part (b) into mod-arg form first and then dividing by z^3 but it didn't work.
Dividing by $z^3$ before subbing in $z = \cos\theta+i\sin\theta$ seems to be the required approach here.
\begin{align*}
\frac{RHS}{z^3} &= \frac{(z^2+1)(z^2-\sqrt3 z+1)(z^2+\sqrt3 z + 1)}{z^3}\\
&= \left(\frac{z^2+1}{z} \right) \left(\frac{z^2-\sqrt3z+1}{z}\right)\left(\frac{z^2+\sqrt3z+1}{z}\right)\\
&= \left( z +\frac1z\right) \left(z + \frac1z - \sqrt3\right) \left( z + \frac1z + \sqrt3\right).
\end{align*}
When $z = \cos\theta+i\sin\theta$, it can be shown that $z + \frac{1}{z} = 2\cos \theta$. Therefore
$\frac{RHS}{z^3} = (2\cos\theta) \left(2\cos\theta - \sqrt3\right) \left(2\cos\theta + \sqrt3\right).$
You should be able to finish things from here, noting that $\sqrt3 = 2\cos\frac\pi6$.  #### ursa

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« Reply #2476 on: May 31, 2020, 09:12:26 pm »
0
Hello! First time I'm posting on an ATAR Notes forum, so hope I'm doing this right In class, we've just been finishing looking at operations on the complex plane. Long story short, I was having difficulty with a question, and the worked solutions provided by the textbook didn't seem to sufficiently answer the question... I spent the whole lesson working with my teacher and the other kid in my ext. 2 class, but we didn't really come up with anything satisfactory.
The question is:
Prove that the vectors representing the complex numbers u, v and (u - iv)/(1-i) form a right-angled triangle.
Would really appreciate seeing someone's method of solving it! Sorry if my formating is confusing, I couldn't figure out how to attach an image.......
Hope that makes sense, thank you! #### fun_jirachi

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« Reply #2477 on: May 31, 2020, 11:08:43 pm »
+4
Welcome to the forums!

In future, you can use an image hosting service like imgur to upload your picture, then copy a link through to the forum to post up a picture. You've definitely got the post right, don't worry about that I think you seem to have some ideas, so instead of the answer, I'm going to drop a few hints:

Geometric approach
- What does the vector $\frac{u-iv}{1-i}$ actually denote? Can you express it in terms of some other more familiar/simpler vectors?
- Construct a parallelogram
- Consider the properties of a parallelogram and a kite
- Draw a diagram of all of the above!

Algebraic approach
- Consider the arguments of the vectors $u-\frac{u-iv}{1-i}$ and $v-\frac{u-iv}{1-i}$. What do the values of the arguments tell you?

If there's anything else, feel free to ask in this thread Failing everything, but I'm still Flareon up.

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#### vernburn

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« Reply #2478 on: June 28, 2020, 02:53:52 pm »
+1
Hey there!

Consider what happens if you take $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5)$, where $\omega = \cos \frac{2\pi}{9} +i\sin \frac{2\pi}{9}$.

We have that $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = (2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9})$.

We also have that by expanding the brackets and simplifying using part (b), $(\omega + \omega ^8)(\omega ^2 + \omega ^7)(\omega ^4+ \omega ^5) = -1$.

Hence, we have that $(2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1 \implies \cos \frac{2\pi}{9} \cos \frac{4\pi}{9} \cos \frac{\pi}{9} = 1$.

What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to $(2\cos \frac{2\pi}{9})(2\cos \frac{4\pi}{9})(-2\cos \frac{\pi}{9}) = -1$. From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions!

Hope this helps Alternatively, we know that:
\begin{aligned}1+z +z ^2+...+z^8 &= \prod_{k=1}^4\left(z- cis\frac{2k\pi}{9}\right)\left(z-cis\left(-\frac{2k\pi}{9}\right)\right)\\ &=\prod_{k=1}^4\left(z^2-2\cos\frac{2k\pi}{9}z+1\right)\end{aligned}

letting $z=i$:
\begin{aligned}\prod_{k=0}^4\left(-2i\cos\frac{2k\pi}{9}\right)&=1\\ 16\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{2\pi}{3}\cos\frac{8\pi}{9}&=1\\ -\frac{1}{2}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}&=\frac{1}{16}\\ -\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=-\frac{1}{8}\\ \implies \cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}&=\frac{1}{8}\end{aligned}