# Thread: HSC 2017 MX2 Marathon (archive)

1. ## HSC 2017 MX2 Marathon (archive)

Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC.

Q16/Q16+ material to be posted here:

Integration Questions:
http://community.boredofstudies.org/...-marathon.html

Really Extra Curricular/Non-Syllabus:
http://community.boredofstudies.org/...icular-topics/

I encourage all current HSC students in particular to participate in this marathon.

Have fun ^_^

To start off:

$Solve (z - 1)^6+ (z + 1)^6 = 0.$

2. ## Re: HSC 2018 4U Marathon

Originally Posted by si2136
Post any questions within the scope and level of Mathematics Extension 2. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC.

I encourage all current HSC students in particular to participate in this marathon.

Have fun ^_^

To start off:

$Solve (z - 1)^6+ (z + 1)^6 = 0.$
Thread title currently says 2018 marathon. Did you mean 2017 (typo)?

3. ## Re: HSC 2018 4U Marathon

Originally Posted by si2136

To start off:

$Solve (z - 1)^6+ (z + 1)^6 = 0.$
$(z-1)^6 +(z+1)^6 = 2z^6 + 30z^4 + 30z^2 +2 \\
z^6 + 15z^4 + 15z^2 + 2 = 0 = P(z)\\
\text{By inspection} z=\pm i \text{is a root, due to the coefficients being real the conjugate roots theorem applies} \\
\text{Performing Polynomial division of} \frac {P(z)}{z^2+1} \text{we obtain the following:}\\
P(z)=(z^2+1)(z^4+14z^2+1)\\
\text{let } x=z^2\\
z^4 + 14z^2 + 1 = 0\\
\text{Though } x=z^2\\
\therefore z=\pm i, \pm \sqrt{-7 \pm \sqrt{48}}
$

Should I leave the answer in a simpler form?

4. ## Re: HSC 2018 4U Marathon

An alternate method (also a more tedious one, esp at the end):
$\noindent z\neq 1\Rightarrow z-1\neq 0 \\\therefore by dividing all terms by (z-1)^6, we get 1+\frac{(z+1)^6}{(z-1)^6}=0\\\Rightarrow \left ( \frac{z+1}{z-1} \right )^{6}=-1 \\ \Rightarrow \frac{z+1}{z-1}= cis\left ( \frac{k\pi}{3}+\frac{\pi}{6} \right ) where k=0,1,2,3,4,5 \\\Rightarrow \frac{z+1}{z-1}=\pm i, \pm \left ( \frac{\sqrt{3}}{2}+\frac{1}{2}i \right ), \pm \left ( \frac{\sqrt{3}}{2}-\frac{1}{2}i \right )\\ solving for z in all 6 cases, we get our final answer$

5. ## Re: HSC 2018 4U Marathon

How do you go next line in latex without it outputting all the
. I used \\

6. ## Re: HSC 2018 4U Marathon

Originally Posted by frog1944
How do you go next line in latex without it outputting all the
. I used \\
Don't put it on a new line. TeX is single line input.

7. ## Re: HSC 2018 4U Marathon

Originally Posted by InteGrand
Thread title currently says 2018 marathon. Did you mean 2017 (typo)?

8. ## Re: HSC 2017 4U Marathon

How are there two thread titles?

http://prntscr.com/ctafpk

9. ## Re: HSC 2017 4U Marathon

Originally Posted by si2136
How are there two thread titles?

http://prntscr.com/ctafpk
Because the original thread was named 2018 and the name got changed but we didn't reply to the new one.

10. ## Re: HSC 2017 4U Marathon

Are we allowed to write z = r(cos x + i sin x) to z = rcisx

11. ## Re: HSC 2017 4U Marathon

Originally Posted by si2136
Are we allowed to write z = r(cos x + i sin x) to z = rcisx
Yes.

12. ## Re: HSC 2017 4U Marathon

If z=x+iy
Express (z+1)/(z-1) in a+ib form.

13. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
If z=x+iy
Express (z+1)/(z-1) in a+ib form.
$\frac{x^2 + y^2 -1}{x^2 + y^2 +1 -2x} -2i \cdot \frac{y}{x^2 + y^2 +1 -2x}$

Ceebs typing the working out... But hint : Don't sub z in straight away.

14. ## Re: HSC 2017 4U Marathon

Originally Posted by Drsoccerball
$\frac{x^2 + y^2 -1}{x^2 + y^2 +1 -2x} -2i \cdot \frac{y}{x^2 + y^2 +1 -2x}$

Ceebs typing the working out... But hint : Don't sub z in straight away.
Why do you recommend to not sub z in straight away. Usually I do.

15. ## Re: HSC 2017 4U Marathon

Originally Posted by si2136
Why do you recommend to not sub z in straight away. Usually I do.
Probs cause realising the denominator becomes less tedious

16. ## Re: HSC 2017 4U Marathon

This is much easier to do. Try subbing in the terms when you have 3 or more terms gg...
$\frac{z + 1}{z-1} \cdot \frac{\bar{z} -1}{\bar{z} - 1}$

17. ## Re: HSC 2017 4U Marathon

$\text{Let }z,w\in \mathbb{C}$

$\\\text{a) Prove that }|Re(z\overline{w})|<|z||w|\\ \text{b) Simplify }z\overline{z}\\ \text{c) Hence prove the triangle inequality }|z+w| \le |z|+|w|$

18. ## Re: HSC 2017 4U Marathon

${Solve}, (z-1)^6 + (z+1)^6 = 0$

19. ## Re: HSC 2017 4U Marathon

Originally Posted by Rathin
If z=x+iy
Express (z+1)/(z-1) in a+ib form.
(2x-1)+2iy

20. ## Re: HSC 2017 4U Marathon

Originally Posted by Kingom
(2x-1)+2iy
Did you assume |z| = 1 ?

21. ## Re: HSC 2017 4U Marathon

Originally Posted by Drsoccerball
Did you assume |z| = 1 ?
no i just multiplied by z+1/z+1

22. ## Re: HSC 2017 4U Marathon

$\frac{z + 1}{z-1} \cdot \frac{\bar{z} -1}{\bar{z} - 1}$

$\frac{z\bar{z} - z + \bar{z} - 1}{z\bar{z} -z - \bar{z} + 1}$

$Take note that : z\bar{z} = x^2 + y^2$

$\frac{x^2 + y^2 - 1 - 2iy }{x^2 + y^2 -2x + 1}$

$\frac{x^2 + y^2 -1}{x^2 + y^2 +1 -2x} -2i \cdot \frac{y}{x^2 + y^2 +1 -2x}$

23. ## Re: HSC 2017 4U Marathon

Originally Posted by leehuan
$\text{Let }z,w\in \mathbb{C}$

$\\\text{a) Prove that }|Re(z\overline{w})|<|z||w|\\ \text{b) Simplify }z\overline{z}\\ \text{c) Hence prove the triangle inequality }|z+w| \le |z|+|w|$
Ill get the ball rolling... THE SOCCERBALL ROLLING!!! YOU GET IT?????! pls save me from finals...
i) $Let z = x + iy and w = a + ib$

$By inspection notice that: (ay - bx)^2 > 0$

$a^2y^2 -2abxy +b^2x^2 > 0$

$a^2x^2 +a^2y^2 +b^2x^2 +b^2y^2 > a^2x^2 +2abxy +b^2y^2$

$(a^2 + b^2)(x^2 + y^2) > (ax+by)^2$

$|z|^2|w|^2 > (Re(z\bar{w}))^2$

$|z||w| > |Re(z\bar{w})|$

24. ## Re: HSC 2017 4U Marathon

$\text{a) Let }z=r_1(\cos \theta_1+i\sin \theta_1), w=r_2(\cos \theta_2+i\sin \theta_2)$

\begin{align*}Re(z\overline{w})&=Re(r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+ \theta_2)))\\ &= r_1r_2(\cos (\theta_1+\theta_2))\\ &\le r_1r_2\\ &=|z||w|\end{align*}

25. ## Re: HSC 2017 4U Marathon

$\noindent Or alternatively, for any complex number \zeta, |\Re \left(\zeta\right)| \leq |\zeta|, because |\zeta|^2 = \left( \Re \left(\zeta\right) \right)^2 + \left(\Im \left(\zeta\right)\right)^2 \geq \left(\Re \left(\zeta\right) \right)^2 \Rightarrow |\zeta| \geq \Re \left(\zeta\right). This implies the result using properties of modulus of a product and of a conjugate.$

$\noindent Geometrically, |\zeta| is the hypotenuse of a right triangle with |\Re \left(\zeta\right)| as another leg, so of course |\zeta| \geq |\Re \left(\zeta\right)|.$

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