HSC 2017 MX2 Marathon (archive) (3 Viewers)

si2136 · Oct 13, 2016

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:
http://community.boredofstudies.org...-2/360463/hsc-2017-mx2-marathon-advanced.html

Integration Questions:
http://community.boredofstudies.org...355439/hsc-2017-mx2-integration-marathon.html

Really Extra Curricular/Non-Syllabus:
http://community.boredofstudies.org/238/extracurricular-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.$

InteGrand · Oct 13, 2016

Re: HSC 2018 4U Marathon

si2136 said:
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)?

frog1944 · Oct 13, 2016

Re: HSC 2018 4U Marathon

si2136 said:
To start off:

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

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

jathu123 · Oct 13, 2016

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$

frog1944 · Oct 13, 2016

Re: HSC 2018 4U Marathon

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

Paradoxica · Oct 13, 2016

Re: HSC 2018 4U Marathon

frog1944 said:
How do you go next line in latex without it outputting all the <br/>. I used \\

Don't put it on a new line. TeX is single line input.

Drsoccerball · Oct 13, 2016

Re: HSC 2018 4U Marathon

InteGrand said:
Thread title currently says 2018 marathon. Did you mean 2017 (typo)?

Well there goes this thread...

si2136 · Oct 13, 2016

Re: HSC 2017 4U Marathon

How are there two thread titles?

http://prntscr.com/ctafpk

Drsoccerball · Oct 13, 2016

Re: HSC 2017 4U Marathon

si2136 said:
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.

si2136 · Oct 15, 2016

Re: HSC 2017 4U Marathon

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

Drsoccerball · Oct 16, 2016

Re: HSC 2017 4U Marathon

si2136 said:
Are we allowed to write z = r(cos x + i sin x) to z = rcisx

Yes.

Green Yoda · Oct 16, 2016

Re: HSC 2017 4U Marathon

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

Drsoccerball · Oct 16, 2016

Re: HSC 2017 4U Marathon

Rathin said:
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.

si2136 · Oct 16, 2016

Re: HSC 2017 4U Marathon

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

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.

DatAtarLyfe · Oct 16, 2016

Re: HSC 2017 4U Marathon

si2136 said:
Why do you recommend to not sub z in straight away. Usually I do.

Probs cause realising the denominator becomes less tedious

Drsoccerball · Oct 16, 2016

Re: HSC 2017 4U Marathon

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

leehuan · Oct 16, 2016

Re: HSC 2017 4U Marathon

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

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

si2136 · Oct 17, 2016

Re: HSC 2017 4U Marathon

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

Kingom · Oct 17, 2016

Re: HSC 2017 4U Marathon

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

(2x-1)+2iy

Drsoccerball · Oct 17, 2016

Re: HSC 2017 4U Marathon

Kingom said:
(2x-1)+2iy

Did you assume |z| = 1 ?

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