MX2 Marathon (2 Viewers)

altSwift · Mar 3, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$\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 said:
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.

$\noindent Hint: Draw the loci on an argand diagram and consider circle geometry theorems$

CapitalSwine · Mar 13, 2018

Re: HSC 2018 MX2 Marathon

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

Paradoxica · Mar 13, 2018

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 · Mar 18, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
$\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$

i)
$\begin{aligned} z\alpha - \bar{z}\bar{\alpha} &= 0 \\ z\alpha &= \overline{z\alpha} \\ \mathrm{Im}(z\alpha) &= 0\end{aligned}$

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

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

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

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

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

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

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

Simplify and we get the quadratic equation

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

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

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

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

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

Using equation $(1)$ ,

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

Therefore,

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

___________________________________________________

Considering the case where $a = 0$ :

$0(y) + bx = 0$

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

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

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

altSwift · Mar 22, 2018

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 · Mar 22, 2018

Re: HSC 2018 MX2 Marathon

Yep, it's broken for me as well.

pikachu975 · Mar 22, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
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 said:
Yep, it's broken for me as well.

Thanks for notifying us!

altSwift · Mar 23, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
$\begin{aligned} |z_1||z_2| &= \sqrt{x_1\,^2+\frac{b^2}{a^2}x_1\,^2} \times \sqrt{x_2\,^2+\frac{b^2}{a^2}x_2\,^2} \\ &= x_1x_2 \sqrt{1 + 2\frac{b^2}{a^2}+\frac{b^4}{a^4}}\\ &= x_1x_2\left(1+\frac{b^2}{a^2}\right) \\ &= x_1x_2\left(\frac{a^2+b^2}{a^2}\right) \end{aligned}$

where did you get that part from? nice solution btw

Paradoxica · Mar 23, 2018

Re: HSC 2018 MX2 Marathon

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

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

fan96 · Mar 24, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
where did you get that part from? nice solution btw

$z_1 = x_1 + iy_1$

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

$y = -\frac b a x$

which gives us

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

therefore

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

fan96 · Mar 24, 2018

Re: HSC 2018 MX2 Marathon

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

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

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

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

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

Equating real and imaginary parts gives us

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

Squaring,

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

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

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

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

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

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

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

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

So cleaning up, we have

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

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

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

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

Paradoxica · Mar 24, 2018

Re: HSC 2018 MX2 Marathon

fan96 said:
Let $(a+ib)^2 = x+iy$ .

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

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

Equating real and imaginary parts gives us

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

Squaring,

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

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

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

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

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

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

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

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

So cleaning up, we have

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

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

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

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

$\sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)$

fan96 · Mar 24, 2018

Re: HSC 2018 MX2 Marathon

Paradoxica said:
$\sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)$

Woops. Forgot to rationalise the denominator.

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

altSwift · Apr 1, 2018

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

InteGrand · Apr 1, 2018

Re: HSC 2018 MX2 Marathon

altSwift said:
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 · Apr 15, 2018

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 · Apr 15, 2018

Re: HSC 2018 MX2 Marathon

CapitalSwine said:
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 · Apr 15, 2018

Re: HSC 2018 MX2 Marathon

CapitalSwine said:
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 $P(x) =x^3-x^2-24x$ .

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

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

Paradoxica · Apr 29, 2018

Re: HSC 2018 MX2 Marathon

mrbunton said:
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

CapitalSwine · May 28, 2018

Re: HSC 2018 MX2 Marathon

Cant seem to figure out how to do this question:
https://imgur.com/a/jRaWrmq

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