# Thread: HSC 2018 MX2 Marathon

1. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
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?

2. ## 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.

3. ## Re: HSC 2018 MX2 Marathon

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

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

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

4. ## Re: HSC 2018 MX2 Marathon

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

5. ## 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

6. ## Re: HSC 2018 MX2 Marathon

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

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

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

7. ## 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?

8. ## Re: HSC 2018 MX2 Marathon

Yep, it's broken for me as well.

9. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
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?
Originally Posted by fan96
Yep, it's broken for me as well.
Thanks for notifying us!

10. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96

\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

11. ## 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

12. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
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

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

13. ## 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
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.

14. ## Re: HSC 2018 MX2 Marathon

Originally Posted by fan96
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.
$\sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)$

15. ## Re: HSC 2018 MX2 Marathon

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

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

16. ## Re: HSC 2018 MX2 Marathon

No idea how to do ii):

capture.PNG

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

17. ## Re: HSC 2018 MX2 Marathon

Originally Posted by altSwift
No idea how to do ii):

capture.PNG

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.

18. ## 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.

19. ## Re: HSC 2018 MX2 Marathon

Originally Posted by CapitalSwine
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

20. ## Re: HSC 2018 MX2 Marathon

Originally Posted by CapitalSwine
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.

21. ## Re: HSC 2018 MX2 Marathon

For which a ∈ R is the sum of the squares of the zeroes of x for
x^2 − (a − 2)x − a − 1 minimal

22. ## Re: HSC 2018 MX2 Marathon

Originally Posted by mrbunton
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

23. ## Re: HSC 2018 MX2 Marathon

P(√x)=0 has roots α², β²

clearly the minimum is 5, for a=1
correct

24. ## Re: HSC 2018 MX2 Marathon

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

25. ## Re: HSC 2018 MX2 Marathon

RPQ and PQB are supplementary (parallel lines)

PQB = QSB (angle in alt segment)

hence RPQ + PSB = 180

since opposite angles are supplementary, RSQP is cyclic?

been 4 years hope I am right :P

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