MX2 Marathon (2 Viewers)

fan96 · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

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

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

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

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

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

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

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

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

altSwift · Feb 17, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
$$Let $z = (\sqrt{3} + i)^m$ where $m \in \mathbb{Z}^+$

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

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

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

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

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

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

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

Correct!
Here is another (I really like these questions)

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

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

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

fan96 · Feb 18, 2018

Re: HSC 2018 MX2 Marathon

The equation does not seem to hold for $n=3$ .

https://www.wolframalpha.com/input/?i=(1+sqrt3)^(2*3)+(1-sqrt3)^(2*3)=2^(2(3)+1)

Is there supposed to be an $i$ somewhere?

jathu123 · Feb 18, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
The equation does not seem to hold for $n=3$ .

https://www.wolframalpha.com/input/?i=(1+sqrt3)^(2*3)+(1-sqrt3)^(2*3)=2^(2(3)+1)

Is there supposed to be an $i$ somewhere?

Yes it should be $(1 + \sqrt3 i)^{2n} + (1 - \sqrt3 i)^{2n} = 2^{2n + 1}$

fan96 · Feb 18, 2018

Re: HSC 2018 MX2 Marathon

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

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

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

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

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

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

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

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

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

If $3 \mid n$ , then

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

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

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

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

altSwift · Feb 18, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
The equation does not seem to hold for $n=3$ .

https://www.wolframalpha.com/input/?i=(1+sqrt3)^(2*3)+(1-sqrt3)^(2*3)=2^(2(3)+1)

Is there supposed to be an $i$ somewhere?

jathu123 said:
Yes it should be $(1 + \sqrt3 i)^{2n} + (1 - \sqrt3 i)^{2n} = 2^{2n + 1}$

My bad, yeah thats the correct equation

fan96 · Feb 19, 2018

Re: HSC 2018 MX2 Marathon

An interesting roots of unity question from the SGS notes.

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

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

altSwift · Feb 19, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
An interesting roots of unity question from the SGS notes.

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

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

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

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

$$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})$

$\noindent$Using 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)$$

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

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

altSwift · Feb 19, 2018

Re: HSC 2018 MX2 Marathon

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

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

sharky564 · Feb 19, 2018

Re: HSC 2018 MX2 Marathon

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

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

Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$ .

altSwift · Feb 20, 2018

Re: HSC 2018 MX2 Marathon

sharky564 said:
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$ .

Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too

altSwift · Feb 20, 2018

Re: HSC 2018 MX2 Marathon

sharky564 said:
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$ .

altSwift said:
Yeah lol, I was trying to force the weird perfect square thing that cambridge does. But that works too

$For reference anyway:$

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

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

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

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

$Factorise as a quadratic from there$

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

fan96 · Feb 20, 2018

Re: HSC 2018 MX2 Marathon

sharky564 said:
Trivially, $\noindent P(x)=(2x^2+3x-2)(x^2+2x-1)=(x+2)(2x-1)(x+1-\sqrt{2})(x+1+\sqrt{2})$ .

What made the polynomial so easy to factor?

fan96 · Feb 23, 2018

Re: HSC 2018 MX2 Marathon

Note that only even numbers can be halved to obtain an integer, and $odd + odd = even$ and $even + even= even$ . 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 $5\times 5 + 4 \times 5 = 45$

fan96 · Feb 23, 2018

Re: HSC 2018 MX2 Marathon

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

mrbunton · Feb 24, 2018

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

altSwift · Feb 25, 2018

Re: HSC 2018 MX2 Marathon

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

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

$$\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$

mrbunton · Feb 28, 2018

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.

altSwift · Feb 28, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
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 · Feb 28, 2018

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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