HSC 2016 MX2 Complex Numbers Marathon (archive) (3 Viewers)

math man · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

A kites diagonals do not bisect each other

parad0xica · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
A kites diagonals do not bisect each other

Whoops, I meant that the chord $U_1U_2$ is bisected

Source: http://www.coolmath.com/reference/kites

math man · Apr 17, 2016

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 · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
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 · Apr 17, 2016

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 · Apr 17, 2016

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 · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
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.

parad0xica · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
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

$\overline{w}(u_1+u_2) + w(\overline{u_1}+\overline{u_2}) =4$

Taking reals

$Re(\overline{w}(u_1+u_2))= 2$

Now finding imaginary

$\overline{w}(u_1+u_2) - w(\overline{u_1}+\overline{u_2}) = c(\overline{u_1+u_2})(u_1+u_2) - c(u_1+u_2)(\overline{u_1+u_2}) = 0$

$\therefore 2Im(\overline{w} (u_1+u_2)) = 0$

$\overline{w}(u_1+u_2) = 2 + 0i$

$|w||u_1+u_2| = |\overline{w}||u_1+u_2 |=|\overline{w}(u_1+u_2) |= 2$

Of course,

$$Radius $ = \frac{|w|}{2} = |u_1+u_2|^{-1}$

How's this?

__________________________

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

math man · Apr 17, 2016

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 · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

parad0xica said:
Adding

$\overline{w}(u_1+u_2) + w(\overline{u_1}+\overline{u_2}) =4$

Taking reals

$Re(\overline{w}(u_1+u_2))= 2$

Now finding imaginary

$\overline{w}(u_1+u_2) - w(\overline{u_1}+\overline{u_2}) = c(\overline{u_1+u_2})(u_1+u_2) - c(u_1+u_2)(\overline{u_1+u_2}) = 0$

$\therefore 2Im(\overline{w} (u_1+u_2)) = 0$

$\overline{w}(u_1+u_2) = 2 + 0i$

$|w||u_1+u_2| = |\overline{w}||u_1+u_2 |=|\overline{w}(u_1+u_2) |= 2$

Of course,

$$Radius $ = \frac{|w|}{2} = |u_1+u_2|^{-1}$

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 · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

KingOfActing said:
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 · Apr 17, 2016

Re: HSC 2016 Complex Numbers Marathon

KingOfActing said:
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 · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

NEW
Q: sketch arg(z-2)=arg(z+i)

KingOfActing · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

math man said:
NEW
Q: sketch arg(z-2)=arg(z+i)

It is the line $y = \frac{1}{2}x - 1$ with the domain restricted to $x \in \mathbb{R} \setminus [0, 2]$ 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.

KingOfActing · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

$$Using complex numbers, prove that $ \tan^{-1}{1} + \tan^{-1}{2} + \tan^{-1}{3} = \pi$

seanieg89 · Apr 18, 2016

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?

tywebb · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

KingOfActing said:
$$Using complex numbers, prove that $ \tan^{-1}{1} + \tan^{-1}{2} + \tan^{-1}{3} = \pi$

Better still, prove that

$x>\textstyle\sqrt{y^2+1}\Rightarrow\tan^{-1}(x-y)+\tan^{-1}(x+y)+\tan^{-1}\frac{2x}{x^2-y^2-1}=\pi$

and hence if x=2 and y=1 then

$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$

KingOfActing · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

seanieg89 said:
By using complex numbers, show that for a quadrilateral ABCD, we have:

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

When does equality occur?

$Letting the complex number $a$ denote the point $A$, and so on:$
$\begin{align*}AB \cdot CD + AD \cdot BC &= |a-b||c-d| + |a-d||b-c| \\&\geq \left| (a-b)(c-d) + (a-d)(b-c)\right|\\&=|ac-ad-bc+bd+ab-bd-ac+dc|\\&=|a(b-d) - c(b-d)|\\&=|a-c||b-d| \\&= AC \cdot BD\\\text{Q.E.D.}\end{align*}$
$The inequality comes from the triangle inequality $|x| + |y| \geq |x+y|$, which can be proved by drawing a triangle with lengths $x$ and $y$. Equality occurs only when $\arg{x} = \arg{y}$ and hence equality occurs when $\arg\left((a-b)(c-d)\right) = \arg\left( (a-d)(b-c)\right)$. This can be simplified down to $\arg(a-b) - \arg(c-b) - \pi = -\left(\arg(c-d) -\arg(a-d)\right) \Longleftrightarrow \angle ABC + \angle CDA = \pi$. Hence equality is achieved if and only if the quadrilateral is cyclic.$

parad0xica · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

KingOfActing said:
$$Using complex numbers, prove that $ \tan^{-1}{1} + \tan^{-1}{2} + \tan^{-1}{3} = \pi$

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"

Paradoxica · Apr 18, 2016

Re: HSC 2016 Complex Numbers Marathon

tywebb said:
Better still, prove that

$x>\textstyle\sqrt{y^2+1}\Rightarrow\tan^{-1}(x-y)+\tan^{-1}(x+y)+\tan^{-1}\frac{2x}{x^2-y^2-1}=\pi$

and hence if x=2 and y=1 then

$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$

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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