# 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

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