1. Re: HSC 2018 MX2 Marathon

Originally Posted by pikachu975
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!!

2. Re: HSC 2018 MX2 Marathon

Originally Posted by pikachu975
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

3. Re: HSC 2018 MX2 Marathon

Originally Posted by sssona09
how about if re(z) is 2Im(z) and z^2-4i is real

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

4. Re: HSC 2018 MX2 Marathon

Prove the ellipses:

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

where k is an arbitrary real number

have the same area as the ellipse

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

5. Re: HSC 2018 MX2 Marathon

Prove the ellipses:

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

have the same area as the ellipse

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

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

has area

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

6. Re: HSC 2018 MX2 Marathon

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

7. 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

9. Re: HSC 2018 MX2 Marathon

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

10. Re: HSC 2018 MX2 Marathon

Could show me some working pls. Sorry for the trouble

12. Re: HSC 2018 MX2 Marathon

This one pls
Capture 3.JPG

13. Re: HSC 2018 MX2 Marathon

Originally Posted by Pakka
This one pls
Capture 3.JPG
Realise the denominator and use De Moivre's at the end to get the n in 2ntheta

14. 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)

15. Re: HSC 2018 MX2 Marathon

Originally Posted by CapitalSwine
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.

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