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HSC 2014 MX2 Marathon (archive) (6 Viewers)

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Sy123

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Re: HSC 2014 4U Marathon

Facedesk*
I just recapped, I kept trying to think visually but i guess algebraic would be best...
Its just letting z=x+i that threw me off.
So i let z=x+iy.
for i) it is clear that locus is y=1 by equating coeff
ii) Similarly from your expansion y=1 should also be locus for z squared.
How do you let z=x+iy?

Think of it this way

since z=(x^2-1) + 2x i

What is z if x=1? or x=2? or x=3?

You get a series of dots, but they follow a certain curve, this curve is the locus. How do you find this locus given z?
 

Ikki

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Re: HSC 2014 4U Marathon

How do you let z=x+iy?

Think of it this way

since z=(x^2-1) + 2x i

What is z if x=1? or x=2? or x=3?

You get a series of dots, but they follow a certain curve, this curve is the locus. How do you find this locus given z?
If x=1: z^2=2i, If x=2:z^2=3+4i, If x=3: z^2=8+6i

If i let z=x+iy but also x+i
then x^2-y^2+2xyi=x^2-1+2xi
Equating imaginaries, y=1 right? But clearly not the case by substituting...

EDIT: We could just take them as points (0,2) (3,4) and (8,6). SYYYYYYYYYYYYYY Mind is blowing here
 
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Sy123

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Re: HSC 2014 4U Marathon

If x=1: z^2=2i, If x=2:z^2=3+4i, If x=3: z^2=8+6i

If i let z=x+iy but also x+i
then x^2-y^2+2xyi=x^2-1+2xi
Equating imaginaries, y=1 right? But clearly not the case by substituting...

EDIT: We could just take them as points (0,2) (3,4) and (8,6). SYYYYYYYYYYYYYY Mind is blowing here
But remember y=/=1 here!

its not z=x+i, its z=(x+i)^2 which is completley different

just imagine its z=t + i instead of x + i just so its not confusing

so z= x+iy = (t+i)^2 = (t^2-1) + 2t i

Hence x=(t^2-1) , y= 2t

t = y/2, x= (y^2/4-1)

Hence y^2 = 4(x+1)

Therefore the locus is the equation y^2 = 4(x+1) in the cartesian plane.
 

Trebla

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Re: HSC 2014 4U Marathon

Not sure if that is within the HSC syllabus?
 

Ikki

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Re: HSC 2014 4U Marathon

Ah like parametrics aye, yeh the t thing helped alot
TREBLAAA hi :)
 

Sy123

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Re: HSC 2014 4U Marathon

Not sure if that is within the HSC syllabus?
Oh I was under the impression it was
(still a good problem though it was not useless)
 

Sy123

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Re: HSC 2014 4U Marathon

(Definitely within syllabus)

 

Ikki

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Re: HSC 2014 4U Marathon

(Definitely within syllabus)

i) Hmm with this one i simply drew up the unit circle and placed z on the argand diagram between 0 and pi/2.
Similarly, for z^2 the domain extends to pi so I placed z^2 in the second quadrant which also lies in the unit circle.
By the modulus of z is clearly 1 for both numbers and similarly the distance from z and z^2 to (z+z^2)=1 via vector addition.
Therefore all sides of the quadrilateral (0,z,z^2,z+z^2) are equal, therefore it can be classified as a rhombus.

ii) let theta=x for my convienience.
z=cisx
z^2=cis2x (By demoivres)
z+z^2=cisx+cis2x= {cosx + cos2x} + i(sinx+sin2x)

But z^2 also = cos^2x-sin^2x + 2sinxcosxi
Therefore, z+z^2={cos^2x-sin^2x}+cosx+2sinxcosxi

Now observing LHS of what we need to show, we equate the Reals:

ANDDD This is where i'm not sure, am I on the right track?
 
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Sy123

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Re: HSC 2014 4U Marathon

i) Hmm with this one i simply drew up the unit circle and placed z on the argand diagram between 0 and pi/2.
Similarly, for z^2 the domain extends to pi so I placed z^2 in the second quadrant which also lies in the unit circle.
By the modulus of z is clearly 1 for both numbers and similarly the distance from z and z^2 to (z+z^2)=1 via vector addition.
Therefore all sides of the quadrilateral (0,z,z^2,z+z^2) are equal, therefore it can be classified as a rhombus.

ii) let theta=x for my convienience.
z=cisx
z^2=cis2x (By demoivres)
z+z^2=cisx+cis2x= {cosx + cos2x} + i(sinx+sin2x)

But z^2 also = cos^2x-sin^2x + 2sinxcosxi
Therefore, z+z^2={cos^2x-sin^2x}+cosx+2sinxcosxi

Now observing LHS of what we need to show, we equate the Reals:

ANDDD This is where i'm not sure, am I on the right track?
The first is correct, for the second one, try finding (z+z^2) in modulus argument form by considering the vector geometrically (use the first part)
 

Sy123

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Re: HSC 2014 4U Marathon

 

dunjaaa

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Re: HSC 2014 4U Marathon

Domain: x=±1
Range: y=±pi
I drew each separate graph and they shared the common points of (1,pi/2),(-1,-pi/2) and hence by the summing the graphs, y-ordinate doubles. Everywhere else is undefined, you get 2 dots at (x=1,y=pi) and (x=-1,y=-pi)
 

Sy123

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Re: HSC 2014 4U Marathon

 

hit patel

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Re: HSC 2014 4U Marathon

Volumes Question :
Volumes.jpg
 

Sy123

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Re: HSC 2014 4U Marathon

I am getting I_(n-2) + I_n = 1/(2n-1) . Why so different?
If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get

I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1)

I_n + I_(n+1) = 1/(2n+1)
 

hit patel

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Re: HSC 2014 4U Marathon

If you mean I_(n-1) + I_n = 1/(2n-1) then this is fine since if we just make n -> n+1 we get

I_((n+1) - 1) + I_(n+1) = 1/(2(n+1)-1)

I_n + I_(n+1) = 1/(2n+1)
Sorry its the same thing. Didnt see that. TOo much caffeine.
 
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