MX2 Marathon (2 Viewers)

sssona09

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Re: HSC 2018 MX2 Marathon

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

thank you
 

sssona09

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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

I cant get the answer.. the answer is +- (2+i)
 

Paradoxica

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Re: HSC 2018 MX2 Marathon

Prove the ellipses:



where k is an arbitrary real number

have the same area as the ellipse

 
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Paradoxica

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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

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Re: HSC 2018 MX2 Marathon

Could show me some working pls. Sorry for the trouble
 

Pakka

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Re: HSC 2018 MX2 Marathon

Never mind, I got it. Cheers, the advice made things easier.
 

CapitalSwine

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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

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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

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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.

Sent from my Redmi Note 4 using Tapatalk
 

altSwift

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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...
 

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