MX2 Marathon (1 Viewer)

sssona09

Member
Re: HSC 2018 MX2 Marathon

Prove by Induction that $\bg_white \overline{z_1+z_2+\dots + z_n} = \overline{z}_1 +\overline{z}_2 + \dots \overline{z}_n$
is this the question?
yes

sssona09

Member
Re: HSC 2018 MX2 Marathon

find z suvh that (Im)z=2 and z^2 is real

thank you

pikachu975

Re: HSC 2018 MX2 Marathon

find z suvh that (Im)z=2 and z^2 is real

thank you
So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i

sssona09

Member
Re: HSC 2018 MX2 Marathon

So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i
thank you Pikachu!!

sssona09

Member
Re: HSC 2018 MX2 Marathon

So let z = x + 2i

(x+2i)^2 = x^2 + 4xi - 4
Since it is real then Im(z^2) = 0
so 4x = 0
x = 0

Hence z = 2i
how about if re(z) is 2Im(z) and z^2-4i is real

1729

Active Member
Re: HSC 2018 MX2 Marathon

how about if re(z) is 2Im(z) and z^2-4i is real

$\bg_white \noindent Let z=x+yi, then \Re(z)=2\Im(z) \implies x=2y. Now z = 2y + yi \implies z^2 - 4i = 3y^2+(4y^2-4)i. Being real, 4y^2-4=0\implies y=\pm 1 \implies x = \pm 2. Thus, z=\pm (2+i)$

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-insert title here-
Re: HSC 2018 MX2 Marathon

Prove the ellipses:

$\bg_white \frac{(x + k y)^2}{a^2} + \frac{y^2}{b^2} = 1, \frac{x^2}{a^2} + \frac{(kx + y)^2}{b^2} = 1$

where k is an arbitrary real number

have the same area as the ellipse

$\bg_white \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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Kingom

Member
Re: HSC 2018 MX2 Marathon

Prove the ellipses:

$\bg_white \frac{(x + k y)^2}{a^2} + \frac{y^2}{b^2} = 1, \frac{x^2}{a^2} + \frac{(kx + y)^2}{b^2} = 1$

have the same area as the ellipse

$\bg_white \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Continuing on, prove that the area of the ellipse defined by

$\bg_white \frac{(a x + b y)^2}{p^2} + \frac{(cx + dy)^2}{q^2} = 1$

has area

$\bg_white \frac{\pi p q}{|ad - bc|}$

Sy123

This too shall pass
Re: HSC 2018 MX2 Marathon

$\bg_white \\ Referring to the diagram above, where the point \ P \ is the intersection of curves \ y =x, \ y = \cos x, which region is larger in area, A_1 or A_2 ? Prove your answer without any use of a calculator (except for very basic facts like that \pi > 3 and so on, another exercise is to prove that \pi > 3)$

-insert title here-
Re: HSC 2018 MX2 Marathon

Letting the value of the solution to x = cos(x) be r, the following area expressions are obtained through integration:

A₁ = sin(r) - r²/2

A₂ = 1 - sin(r) + r²/2

A₂ - A₁ = 1 - 2sin(r) + r²

r > sin(r) (proof is trivial and left as an exercise)

- 2sin(r) > - 2r

1 - 2sin(r) + r² > 1 - 2r + r² = (1-r)² > 0

A₂ - A₁ > 0

A₂ > A₁

Pakka

New Member
Re: HSC 2018 MX2 Marathon

These pls

sida1049

Well-Known Member
Re: HSC 2018 MX2 Marathon

Change to polar form, apply de Moivre's theorem, done. For all of them.

Pakka

New Member
Re: HSC 2018 MX2 Marathon

Could show me some working pls. Sorry for the trouble

Pakka

New Member
Re: HSC 2018 MX2 Marathon

Pakka

New Member
Re: HSC 2018 MX2 Marathon

This one pls

CapitalSwine

New Member
Re: HSC 2018 MX2 Marathon

Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)

1729

Active Member
Re: HSC 2018 MX2 Marathon

Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)
Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.

CapitalSwine

New Member
Re: HSC 2018 MX2 Marathon

Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.
What is the modulus and argument of z in this case? I got an answer that was different to the one in the textbook.

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altSwift

Member
Re: HSC 2018 MX2 Marathon

Find an expression for cos^4x in terms of cos4x and cos2x

Edit: The question is from terry lee so I'm guessing it implies that you use 4U complex techniques rather than 3U, I'm not sure if you can even use 3U for this...