# HSC 2018-2019 MX2 Marathon (1 Viewer)

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

$\bg_white Let z = (\sqrt{3} + i)^m$ where $\bg_white m \in \mathbb{Z}^+$

$\bg_white \therefore z= \left(2\, \text{cis}\, \frac{\pi}{6}\right)^m= 2m \cos \left(\frac{m\pi}{6}\right) + i\left(2m \sin\left(\frac{m\pi}{6}\right)\right)$
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If $\bg_white z$ is real, then $\bg_white Im(z) = 0$

\bg_white \begin{aligned}\therefore 2m \sin\left(\frac{m\pi}{6}\right) &= 0 \\ \sin\left(\frac{m\pi}{6}\right) &= 0 \end{aligned}

The smallest value of $\bg_white m$ is $\bg_white 6$.
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If $\bg_white z$ is purely imaginary, then $\bg_white Re(z) = 0$

\bg_white \begin{aligned}\therefore 2m \cos \left(\frac{m\pi}{6}\right) &= 0 \\ \cos \left(\frac{m\pi}{6}\right) &= 0\end{aligned}

The smallest value of $\bg_white m$ is $\bg_white 3$.

#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

$\bg_white Let z = (\sqrt{3} + i)^m$ where $\bg_white m \in \mathbb{Z}^+$

$\bg_white \therefore z= \left(2\, \text{cis}\, \frac{\pi}{6}\right)^m= 2m \cos \left(\frac{m\pi}{6}\right) + i\left(2m \sin\left(\frac{m\pi}{6}\right)\right)$
_____________________________________________________

If $\bg_white z$ is real, then $\bg_white Im(z) = 0$

\bg_white \begin{aligned}\therefore 2m \sin\left(\frac{m\pi}{6}\right) &= 0 \\ \sin\left(\frac{m\pi}{6}\right) &= 0 \end{aligned}

The smallest value of $\bg_white m$ is $\bg_white 6$.
_____________________________________________________

If $\bg_white z$ is purely imaginary, then $\bg_white Re(z) = 0$

\bg_white \begin{aligned}\therefore 2m \cos \left(\frac{m\pi}{6}\right) &= 0 \\ \cos \left(\frac{m\pi}{6}\right) &= 0\end{aligned}

The smallest value of $\bg_white m$ is $\bg_white 3$.
Correct!
Here is another (I really like these questions)

$\bg_white a) Show that if n is divisible by 3 then:$

$\bg_white (1 + \sqrt{3}i)^{2n} + (1 - \sqrt{3}i)^{2n} = 2^{2n + 1}$

$\bg_white b) Simplify the expression if n is \textbf NOT divisible by 3$

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

$\bg_white (1+i\sqrt{3})^{2n} + (1-i\sqrt{3})^{2n}$

$\bg_white = \left(2 \text{ cis } \frac{\pi}{3} \right)^{2n} + \left(2 \text{ cis } \frac{-\pi}{3} \right)^{2n}$

$\bg_white = 2^{2n}\left( \text{cis } \frac{2n\pi}{3} + \text{cis } \frac{-2n\pi}{3}\right)$

$\bg_white = 2^{2n}\left(\cos\frac{2n\pi}{3} + i \sin \frac{2n\pi}{3} + \cos \frac{-2n\pi}{3} + i \sin \frac{-2n\pi}{3}\right)$

Since $\bg_white \cos x$ is an even function and $\bg_white \sin x$ is an odd function,

$\bg_white = 2^{2n}\left(\cos\frac{2n\pi}{3} + i \sin \frac{2n\pi}{3} + \cos \frac{2n\pi}{3} - i \sin \frac{2n\pi}{3}\right)$

$\bg_white = 2^{2n}\left(2\cos\frac{2n\pi}{3}\right)$

$\bg_white =\left(2^{2n}\times 2^1\right) \left(\cos \frac{2n\pi}{3}\right)$

$\bg_white = 2^{2n+1}\left(\cos \frac{2n\pi}{3}\right)$
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If $\bg_white 3 \mid n$, then

$\bg_white \cos \frac{2n\pi}{3} = \cos 2k\pi, \, k \in \mathbb{Z}$

Which is the general solution for $\bg_white \cos x = 1$

$\bg_white \therefore \cos \frac{2n\pi}{3} = 1$

$\bg_white \therefore 2^{2n+1}\left(\cos \frac{2n\pi}{3}\right) = 2^{2n+1}$

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of $\bg_white -1$ in the form $\bg_white a+ib$.

ii) Hence write $\bg_white z^4+1$ as a product of two quadratic factors with real co-efficients.

#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

i) Find the fourth roots of $\bg_white -1$ in the form $\bg_white a+ib$.

ii) Hence write $\bg_white z^4+1$ as a product of two quadratic factors with real co-efficients.
$\bg_white Roots: cis\frac{\pi}{4} (\frac{1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-\pi}{4} (\frac{1}{\sqrt2} + \frac{-i}{\sqrt2}),$

$\bg_white cis\frac{3\pi}{4} (\frac{-1}{\sqrt2} + \frac{i}{\sqrt2}), cis\frac{-3\pi}{4} (\frac{-1}{\sqrt2} + \frac{-i}{\sqrt2})$

$\bg_white Now z^4 + 1 = (z - cis\frac{\pi}{4})(z - cis\frac{-\pi}{4})(z - cis\frac{3\pi}{4})(z - cis\frac{-3\pi}{4})$

$\bg_white \noindentUsing the fact that z + \bar z = 2\mathrm{Re} (z) and that z\bar z = x^2 + y^2 (and the fact the roots are in conjugate pairs)$

$\bg_white z^4 + 1 = (z^2 - 2(\frac{1}{\sqrt2})z + 1)(z^2 + 2(\frac{1}{\sqrt2})z + 1)$

$\bg_white z^4 + 1 = (z^2 - \sqrt2 z + 1)(z^2 + \sqrt2 z + 1)$

#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$

$\bg_white \noindent P(x) = 0 over C without using long division, sum/product of roots etc or the polynomial remainder theorem$

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#### sharky564

##### Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$

$\bg_white \noindent P(x) = 0 over C without using long division, sum/product of roots etc or the polynomial remainder theorem$
Trivially, $\bg_white \noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.

#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

Trivially, $\bg_white \noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too

#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

Trivially, $\bg_white \noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too
$\bg_white For reference anyway:$

$\bg_white P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2$
$\bg_white = x^2 (2x^2 + 7x + 2 - \frac{7}{x} + \frac{2}{x^2})$

$\bg_white = x^2 \left(2(x^2 + \frac{1}{x^2}) + 7(x - \frac{1}{x}) + 2\right)$

$\bg_white Using (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2$

$\bg_white = x^2 \left(2(x - \frac{1}{x})^2 + 7(x - \frac{1}{x}) + 6\right)$

$\bg_white Factorise as a quadratic from there$

$\bg_white \noindent Idk if there is any point of doing it this way if you can use either the remainder theorem or equate coefficients. Still looked pretty nice when I first saw it$

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

Trivially, $\bg_white \noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$.
What made the polynomial so easy to factor?

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

Note that only even numbers can be halved to obtain an integer, and $\bg_white odd + odd = even$ and $\bg_white even + even= even$. Finally, the average of any two digits is never greater than 9 or less than 0.

If the first digit is odd (1, 3, 5, 7, 9), then the last digit must be odd too (1, 3, 5, 7, 9).

If the first digit is even (2, 4, 6, 8), then the last digit must be even or zero (0, 2, 4, 6, 8).

Any two digits can only have one possible average, so the middle digit doesn't matter.

Therefore the number of digits is $\bg_white 5\times 5 + 4 \times 5 = 45$

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Marathon

Show that the polynomial $\bg_white P(x) = x^3-3x+k$ (where $\bg_white k$ is an integer such that $\bg_white k > 2$) has exactly one real zero $\bg_white r$ where $\bg_white r < -1$.

#### mrbunton

##### Member
Re: HSC 2018 MX2 Marathon

derive equation to get turning points at -1 and 1
find y values of these turning points:
(-1,2+k) and (1,k-2)
since the poly is odd it will have at least one root.
when k>2; both turning points are above the x-axis; so no extra roots are possible.

proving that r<-1 or similar value:
when letting k=2; that poly is less then 0. the poly can be factored into (x+2)(x-1)^2<0
condition is only true at x<-2 or r<-2
-2<-1 so r<-1

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#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

$\bg_white \noindent Given z = x + iy is a variable point and \alpha = a + ib is a fixed point on the Argand plane,$

$\bg_white \noindent (i) Show that z\alpha -\bar{z}\bar{\alpha} = 0 represents a straight line through the origin O$

$\bg_white \noindent (ii) Suppose that z\textsubscript{1} and z\textsubscript{2} are the solutions to the simultaneous equations: z\alpha -\bar{z}\bar{\alpha} = 0 and \mid z - \beta \mid = k , where \beta = p + iq , and where p, q and k are positive real numbers, show that \mid z\textsubscript{1} \mid \mid z\textsubscript{2} \mid = \mid p^2 + q^2 - k^2 \mid$

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#### mrbunton

##### Member
Re: HSC 2018 MX2 Marathon

i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

ii) (x-p)^2+(y-q)^2= k^2
sub y=0 and expand and bring to one side
x^2 -(2p)x +(p^2+ q^2 - k^2) =0
product of roots: z1*z2= p^2 + q^2 - k^2

which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true.

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#### altSwift

##### Member
Re: HSC 2018 MX2 Marathon

i) the line created is bx+ay=0.since you said i) is not the locus of z; and z can be truly anything; then we can observe that bx+ay=0 only yields to purely real values; so it becomes the line y=0 on the imaginary number plane

ii) (x-p)^2+(y-q)^2= k^2
sub y=0 and expand and bring to one side
x^2 -(2p)x +(p^2+ q^2 - k^2) =0
product of roots: z1*z2= p^2 + q^2 - k^2

which is the same thing as we are meant to show. normally one would say the product of the roots are just the real parts of z1 and z2; but there is no imaginary part so condition is still true. But in an exam u would of been expected to spend a couple more lines to actually be rigorous but im lazy.
How can you say that the equation is y=0 on the imaginary plane but then sub into an equation that's on the real number plane? Wouldn't you have to sub in bx + ay = 0?

#### mrbunton

##### Member
Re: HSC 2018 MX2 Marathon

ya rip me. i misinterpreted what z as a variable point meant. i thought it meant z could be anything and that the line was dependent on what z was; although that doesnt make much sense once you think about it.

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