# HSC 2018-2019 MX2 Marathon (1 Viewer)

#### altSwift

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

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

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

$\bg_white \noindent (ii) Suppose that z\textsubscript{1} and z\textsubscript{2} are the solutions to the simultaneous equations: z\alpha -\bar{z}\bar{\alpha} = 0 and \mid z - \beta \mid = k , where \beta = p + iq , and where p, q and k are positive real numbers, show that \mid z\textsubscript{1} \mid \mid z\textsubscript{2} \mid = \mid p^2 + q^2 - k^2 \mid$
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.
$\bg_white \noindent Hint: Draw the loci on an argand diagram and consider circle geometry theorems$

#### CapitalSwine

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

What is the solution of z^2 = i*(conjugate of z)

##### -insert title here-
Re: HSC 2018 MX2 Marathon

x² − y² = y & 2xy = x
x=0, y = -1,0
or
y=½, x = ±½√3

Sent from my iPhone using Tapatalk

#### fan96

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

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

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

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

i.e. $\bg_white z\alpha \in \mathbb{R}$. But,

$\bg_white \mathrm{Im}(z\alpha) = \mathrm{Im}((x+iy)(a+ib)) = bx + ay$

Therefore the locus of $\bg_white z\alpha - \bar{z}\bar{\alpha} = 0$ is the line $\bg_white bx + ay = 0$, which is linear and passes through $\bg_white O$.

The exception is when $\bg_white \alpha = 0$, where the locus of $\bg_white z$ is all complex numbers $\bg_white z \in \mathbb{C}$. Going off the question, I will assume $\bg_white \alpha$ is nonzero.

ii)
\bg_white \begin{aligned}|z-\beta| &= k \\ (x-p)^2+(y-q)^2 &= k^2 \end{aligned}

Using a substitution obtainable from the locus from i), $\bg_white y = -\frac{b}{a}x,\, a \neq 0$, we have

\bg_white \begin{aligned}(x-p)^2+\left(-\frac{b}{a}x-q\right)^2 &= k^2 \\ (x-p)^2+\left(\frac{b}{a}x+q\right)^2 &= k^2 \end{aligned}

Simplify and we get the quadratic equation

$\bg_white \left(1+\frac{b^2}{a^2}\right)x^2 + \left(\frac{2bq}{a}-2p\right)x + (p^2+q^2-k^2) = 0$

Let the roots be $\bg_white x_1, \, x_2$. Then, using the product of roots formula,

$\bg_white x_1x_2 = (p^2+q^2-k^2)\left(\frac{a^2}{a^2+b^2}\right) \noindent\rule{1.5cm}{0.4pt}\, (1)$

Which I will call equation $\bg_white (1)$. Now,

\bg_white \begin{aligned} |z_1||z_2| &= \sqrt{x_1\,^2+\frac{b^2}{a^2}x_1\,^2} \times \sqrt{x_2\,^2+\frac{b^2}{a^2}x_2\,^2} \\ &= x_1x_2 \sqrt{1 + 2\frac{b^2}{a^2}+\frac{b^4}{a^4}}\\ &= x_1x_2\left(1+\frac{b^2}{a^2}\right) \\ &= x_1x_2\left(\frac{a^2+b^2}{a^2}\right) \end{aligned}

Using equation $\bg_white (1)$,

\bg_white \begin{aligned} |z_1||z_2| &= (p^2+q^2-k^2)\left(\frac{a^2}{a^2+b^2}\right)\left(\frac{a^2+b^2}{a^2}\right) \end{aligned}

Therefore,

$\bg_white |z_1||z_2| &= p^2+q^2-k^2 \implies |z_1||z_2| = |p^2+q^2-k^2|$, for $\bg_white a \neq 0$.

___________________________________________________

Considering the case where $\bg_white a = 0$:

$\bg_white 0(y) + bx = 0$

So, either of $\bg_white b$ or $\bg_white x$ or both are zero.

If $\bg_white x = 0, \, z = iy$ and it can be proved in a similar manner that $\bg_white |z_1||z_2|= y_1y_2= p^2+q^2-k^2$.

If $\bg_white b = 0, \, \alpha = 0$, which I addressed in part i).

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#### altSwift

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

I can't load any latex in this thread anymore, I've tried on 3 computers, is anyone else getting this issue or just me?

#### fan96

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

Yep, it's broken for me as well.

#### pikachu975

##### I love trials
Moderator
Re: HSC 2018 MX2 Marathon

I can't load any latex in this thread anymore, I've tried on 3 computers, is anyone else getting this issue or just me?
Yep, it's broken for me as well.
Thanks for notifying us!

#### altSwift

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

\bg_white \begin{aligned} |z_1||z_2| &= \sqrt{x_1\,^2+\frac{b^2}{a^2}x_1\,^2} \times \sqrt{x_2\,^2+\frac{b^2}{a^2}x_2\,^2} \\ &= x_1x_2 \sqrt{1 + 2\frac{b^2}{a^2}+\frac{b^4}{a^4}}\\ &= x_1x_2\left(1+\frac{b^2}{a^2}\right) \\ &= x_1x_2\left(\frac{a^2+b^2}{a^2}\right) \end{aligned}
where did you get that part from? nice solution btw

##### -insert title here-
Re: HSC 2018 MX2 Marathon

Find separate expressions for the real and imaginary parts of $\bg_white \sqrt{x+i y}$

Hence determine $\bg_white \sqrt{x+i y}$ as a sum of independent real and imaginary parts

#### fan96

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

where did you get that part from? nice solution btw
$\bg_white z_1 = x_1 + iy_1$

and $\bg_white z_1$ satisfies the locus in i), so

$\bg_white y = -\frac b a x$

which gives us

$\bg_white z_1 = x_1 - i\frac b a x_1$

therefore

$\bg_white |z_1| = \sqrt{x_1\, ^2 + \frac {b^2} {a^2} x_1\, ^2 }$

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#### fan96

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

Find separate expressions for the real and imaginary parts of $\bg_white \sqrt{x+i y}$

Hence determine $\bg_white \sqrt{x+i y}$ as a sum of independent real and imaginary parts
Let $\bg_white (a+ib)^2 = x+iy$.

(Note that if $\bg_white a = 0$ then $\bg_white y = 0$, $\bg_white x \leq 0$ and we have $\bg_white (ib)^2 = x \implies a + ib = \pm i\sqrt{-x}$. Otherwise,)

Expanding, we get $\bg_white (a^2-b^2) +i(2ab) = x + iy$.

Equating real and imaginary parts gives us

$\bg_white \begin{cases} a^2-b^2 &= x \,\noindent\rule{1cm}{0.4pt} (1) \\ 2ab &= y \,\noindent\rule{1cm}{0.4pt} (2)\end{cases}$

Squaring,

$\bg_white \begin{cases} (a^2-b^2)^2 &= x^2 \\ (2ab)^2 &= y^2 \end{cases}$

Adding and using the identity $\bg_white (A+B)^2 =(A-B)^2 + 4AB$ with $\bg_white A=a^2, \, B=b^2$, we get

$\bg_white (a^2+b^2)^2= x^2+y^2$

Since $\bg_white a^2+b^2$ and $\bg_white x^2+y^2$ are both positive, we can take the positive square roots of both sides to get

$\bg_white a^2+b^2= \sqrt{x^2+y^2}\noindent\rule{1cm}{0.4pt} (3)$

$\bg_white (1) + (3) \implies 2a^2 = x + \sqrt{x^2+y^2} \implies a = \pm \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}$

and since $\bg_white (2) \implies b = \frac{y}{2a},\, a \neq 0$,

$\bg_white b = \pm \frac{y}{2\sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}} = \pm \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}$

So cleaning up, we have

$\bg_white a + ib = \pm \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}\pm i \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}$

and we can get rid of that second $\bg_white \pm$ sign if desired:

$\bg_white a + ib = \pm\left( \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}} + i\,\mathrm{sgn}(y) \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}\right)$

Where $\bg_white \mathrm{sgn}\, x$ is the sign function.

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##### -insert title here-
Re: HSC 2018 MX2 Marathon

Let $\bg_white (a+ib)^2 = x+iy$.

(Note that if $\bg_white a = 0$ then $\bg_white y = 0$, $\bg_white x \leq 0$ and we have $\bg_white (ib)^2 = x \implies a + ib = \pm i\sqrt{-x}$. Otherwise,)

Expanding, we get $\bg_white (a^2-b^2) +i(2ab) = x + iy$.

Equating real and imaginary parts gives us

$\bg_white \begin{cases} a^2-b^2 &= x \,\noindent\rule{1cm}{0.4pt} (1) \\ 2ab &= y \,\noindent\rule{1cm}{0.4pt} (2)\end{cases}$

Squaring,

$\bg_white \begin{cases} (a^2-b^2)^2 &= x^2 \\ (2ab)^2 &= y^2 \end{cases}$

Adding and using the identity $\bg_white (A+B)^2 =(A-B)^2 + 4AB$ with $\bg_white A=a^2, \, B=b^2$, we get

$\bg_white (a^2+b^2)^2= x^2+y^2$

Since $\bg_white a^2+b^2$ and $\bg_white x^2+y^2$ are both positive, we can take the positive square roots of both sides to get

$\bg_white a^2+b^2= \sqrt{x^2+y^2}\noindent\rule{1cm}{0.4pt} (3)$

$\bg_white (1) + (3) \implies 2a^2 = x + \sqrt{x^2+y^2} \implies a = \pm \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}$

and since $\bg_white (2) \implies b = \frac{y}{2a},\, a \neq 0$,

$\bg_white b = \pm \frac{y}{2\sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}} = \pm \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}$

So cleaning up, we have

$\bg_white a + ib = \pm \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}}\pm i \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}$

and we can get rid of that second $\bg_white \pm$ sign if desired:

$\bg_white a + ib = \pm\left( \sqrt{\frac{x + \sqrt{x^2+y^2}}{2}} + i\,\mathrm{sgn}(y) \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}\right)$

Where $\bg_white \mathrm{sgn}\, x$ is the sign function.
$\bg_white \sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)$

#### fan96

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

$\bg_white \sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)$
Woops. Forgot to rationalise the denominator.

$\bg_white \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}}$
$\bg_white = \sqrt{\frac{\frac{1}{2}y^2}{x+\sqrt{x^2+y^2}}} \times \sqrt{\frac{x-\sqrt{x^2+y^2}}{x-\sqrt{x^2+y^2}}} = \sqrt{\frac{\frac{1}{2}y^2(x-\sqrt{x^2+y^2})}{x^2-x^2-y^2}} = \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$

#### altSwift

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

No idea how to do ii):

I know this isn't the best place to put this but i didn't want to dedicate a thread to a 1 mark question that's apparently simple

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#### InteGrand

##### Well-Known Member
Re: HSC 2018 MX2 Marathon

No idea how to do ii):

View attachment 34506

I know this isn't the best place to put this but i didn't want to dedicate a thread to a 1 mark question that's apparently simple
Hint: conjugate root theorem.

#### CapitalSwine

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

I dont really understand why you have to find the stationary points (says so in the answers), would appreciate an expanation on how to solve this problem.

#### pikachu975

##### I love trials
Moderator
Re: HSC 2018 MX2 Marathon

I dont really understand why you have to find the stationary points (says so in the answers), would appreciate an expanation on how to solve this problem.
If you know the stationary points you can get a rough sketch of the graph and hence find the k values (which shift the graph up and down) which satisfy the equation having one real root

#### fan96

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

I dont really understand why you have to find the stationary points (says so in the answers), would appreciate an expanation on how to solve this problem.
For simplicity, let $\bg_white P(x) =x^3-x^2-24x$.

First, sketch the graph of $\bg_white y = P(x)$ (the quickest way to do this is to find the turning points of the graph). The graph of $\bg_white y = P(x)+k$ can be obtained by shifting the graph of $\bg_white y = P(x)$ by $\bg_white k$ units up or down.

Finally, a polynomial has as many real roots as its graph has $\bg_white x$-intercepts. So a polynomial has exactly one real root if and only if it crosses the $\bg_white x$-axis exactly once.

##### -insert title here-
Re: HSC 2018 MX2 Marathon

For which a ∈ R is the sum of the squares of the zeroes of x for
x² − (a − 2)x − a − 1 minimal
P(√x)=0 has roots α², β²

Simplify the equation to form a standard quadratic equation.
x - (a+1) = (a-2)√x

x² - 2(a+1)x + (a+1)² = (a-2)²x

x² - 2(a+1)x + (a+1)² = (a²-4a+4)x

x² - (a²-4a+4+2a+2)x + (a+1)² = 0

x² - (a²-2a+6)x + (a+1)² = 0

α²+β² = a²-2a+6 = (a-1)²+5

clearly the minimum is 5, for a=1

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