HSC 2016 MX2 Complex Numbers Marathon (archive) (2 Viewers)

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math man

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Re: HSC 2016 Complex Numbers Marathon

A kites diagonals do not bisect each other
 

math man

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Re: HSC 2016 Complex Numbers Marathon

Another problem is the centre of the circle is not the midpoint of the chord, it's impossible. I'll give you a hint to solve it really easily in a few seconds. The intersection of the tangents is the point w so if you let OW be the complex number w you can sub z=w and then use the fact w lies on u1 + u2
 

parad0xica

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Re: HSC 2016 Complex Numbers Marathon

Another problem is the centre of the circle is not the midpoint of the chord, it's impossible. I'll give you a hint to solve it really easily in a few seconds. The intersection of the tangents is the point w so if you let OW be the complex number w you can sub z=w and then use the fact w lies on u1 + u2
I can't think of anything for this besides w = c(u1 +u2) where is c is real...
 

math man

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Re: HSC 2016 Complex Numbers Marathon

Sub z=w into each tangent equation since that's the intersection point. Then add each equation and you'll notice if you factorise correctly you get sum of two conjugates
 

math man

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Re: HSC 2016 Complex Numbers Marathon

The best approach to these harder questions is to use the answer and think given what you have what you need to prove, since OW is diameter prove |w|=2/|u1+u2| and a more advanced method is play with that result and you get prove |w(u1+u2)|=2 which looks familiar if you add those two equations I said. I call this the Greek method as you play with the answer and put it in a form that is easier to prove
 

KingOfActing

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Re: HSC 2016 Complex Numbers Marathon

The best approach to these harder questions is to use the answer and think given what you have what you need to prove, since OW is diameter prove |w|=2/|u1+u2| and a more advanced method is play with that result and you get prove |w(u1+u2)|=2 which looks familiar if you add those two equations I said. I call this the Greek method as you play with the answer and put it in a form that is easier to prove
EDIT: I'm just gonna rewrite it without quoting the formula's to see how difficult it is

Edit 2: I rewrote it above, not actually that difficult. More geometrical than complex numbers, but it's still valid.
 
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parad0xica

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Re: HSC 2016 Complex Numbers Marathon

The best approach to these harder questions is to use the answer and think given what you have what you need to prove, since OW is diameter prove |w|=2/|u1+u2| and a more advanced method is play with that result and you get prove |w(u1+u2)|=2 which looks familiar if you add those two equations I said. I call this the Greek method as you play with the answer and put it in a form that is easier to prove
Adding



Taking reals



Now finding imaginary









Of course,



How's this?

__________________________

What you call Greek method is heuristically known as working backwards and thanks for the hints btw
 
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math man

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Re: HSC 2016 Complex Numbers Marathon

That method is great cause you didn't even need the assumption ow lies on sum of u1 and u2 as you followed it using its construction
 

KingOfActing

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Re: HSC 2016 Complex Numbers Marathon

Adding



Taking reals



Now finding imaginary









Of course,



How's this?

__________________________

What you call Greek method is heuristically known as working backwards and thanks for the hints btw
Don't you have to prove that OW is the diameter?
 

InteGrand

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Re: HSC 2016 Complex Numbers Marathon

Don't you have to prove that OW is the diameter?
Yes. Luckily it's pretty quick to prove. Angle OU1W is 90 degrees so by converse of 'angle in a semicircle', OW is the diameter.
 

parad0xica

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Re: HSC 2016 Complex Numbers Marathon

Don't you have to prove that OW is the diameter?
My bad, I forgot this was for the user, wu345.

Angle WU1O is 90 degrees by tangent-radius theorem, hence OW is diameter.

Beaten by the swift puɐɹפǝʇuI
 

math man

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Re: HSC 2016 Complex Numbers Marathon

NEW
Q: sketch arg(z-2)=arg(z+i)
 

KingOfActing

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Re: HSC 2016 Complex Numbers Marathon

NEW
Q: sketch arg(z-2)=arg(z+i)
It is the line with the domain restricted to which can be shown algebraically by using arg(z) - arg(w) = arg(z/w)

Alternatively, the geometrical interpretation is that the angle between the points P(0,-1), Z(x,y) and Q(2,0) is 0.
 

seanieg89

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Re: HSC 2016 Complex Numbers Marathon

By using complex numbers, show that for a quadrilateral ABCD, we have:

AB.CD + AD.BC >= AC.BD.

When does equality occur?
 

parad0xica

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Re: HSC 2016 Complex Numbers Marathon

Highlight below to see answer:


(1+i)(1+2i)(1+3i) = -10

arg((1+i)(1+2i)(1+3i)) = arg(-10 )

arg(1+i)+ arg(1+2i)+arg(1+3i) = \pi

arctan(1) + arctan(2) + arctan(3) = \pi



EDIT: added "spoilers"
 
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Paradoxica

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Re: HSC 2016 Complex Numbers Marathon

Better still, prove that



and hence if x=2 and y=1 then

Actually, I was wondering, do you have any more magical trigonometric identities?

I need them. My life force is running dry...
 
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