Complex (1 Viewer)

nrlwinner · Jan 26, 2010

If A,B represent z1 and z2
OAB is equilateral, prove $(z_1)^2+(z_2)^2=z_1z_2$

What am I doing wrong? Or do I have to do soething with multiplying by Cis pi/3

LHS=

$|(z_1)^2|+|(z_2)^2|$

$|z_1|^2+|z_2|^2$

Let common modulus be z

$|z|^2+|z|^2$

$2|z|^2$

RHS=

$|z_1z_2|$

$|z_1||z_2|$

$|z||z|$

$|z|^2$

untouchablecuz · Jan 26, 2010

$\\z_{1}=z_{2}\times cis(\frac{\pi}{3})\\\frac{z_1}{z_2}=cis(\frac{\pi}{3}) \ (*) $ and $ \frac{z_2}{z_1}=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{z_1}{z_2}+\frac{z_2}{z_1}=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re \left(cis(\frac{\pi}{3})\right)=1\\$Multiply through by $z_1z_2,\\z_1^2+z_2^2=z_1z_2$

this is a nice Q

nrlwinner · Jan 26, 2010

untouchablecuz said:
$\\z_{1}=z_{2}\times cis(\frac{\pi}{3})\\\frac{z_1}{z_2}=cis(\frac{\pi}{3}) \ (*) $ and $ \frac{z_2}{z_1}=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{z_1}{z_2}+\frac{z_2}{z_1}=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re (cis\left(\frac{\pi}{3})\right)=1\\$Multiply through by $z_1z_2,\\z_1^2+z_2^2=z_1z_2$

this is a nice Q

Yeah thanks, but how would you know you had to add.
I tried doing it by subing into the equation, trying to prove LHS=RHS
I got up to.

LHS = $(z_2)^2cis\frac{2\pi }{3}+(z_1)^2cis\frac{-2\pi }{3}$

But cant seem to get it to equal z1z2

untouchablecuz · Jan 26, 2010

gurmies · Jan 27, 2010

untouchablecuz said:
$\\z_{1}=z_{2}\times cis(\frac{\pi}{3})\\\frac{z_1}{z_2}=cis(\frac{\pi}{3}) \ (*) $ and $ \frac{z_2}{z_1}=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{z_1}{z_2}+\frac{z_2}{z_1}=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re \left(cis(\frac{\pi}{3})\right)=1\\$Multiply through by $z_1z_2,\\z_1^2+z_2^2=z_1z_2$

this is a nice Q

This is a nice solution.

nrlwinner · Jan 27, 2010

If
$a+ib=\frac{(x+1)^2}{2x-i}$

Prove

$a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}$

I've been trying to do this for half an hour and I'm getting nowhere.

cyl123 · Jan 27, 2010

check the question... should it be (x+1)^2 instead of (x^2+1)^2

Btw is x a real number? If so.... think modulus

nrlwinner · Jan 27, 2010

Ok. I tried taking modulus and I realised I couldn't get it.

The question was.

If

$a+ib=\frac{(x+i)^2}{2x-i}$

Prove

$a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}$

untouchablecuz · Jan 27, 2010

nrlwinner said:
Ok. I tried taking modulus and I realised I couldn't get it.

The question was.

If

$a+ib=\frac{(x+i)^2}{2x-i}$

Prove

$a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}$

$\\a+ib=\frac{(x+i)^2}{2x-i}\\|a+ib|=\left|\frac{(x+i)^2}{2x-i}\right|=\frac{|(x+i)^2|}{|2x-i|}=\frac{|x+i|^2}{|2x-i|}\\a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}$

nrlwinner · Jan 27, 2010

untouchablecuz said:
$\\a+ib=\frac{(x+i)^2}{2x-i}\\|a+ib|=\left|\frac{(x+i)^2}{2x-i}\right|=\frac{|(x+i)^2|}{|2x-i|}=\frac{|x+i|^2}{|2x-i|}\\a^2+b^2=\frac{(x^2+1)^2}{4x^2+1}$

Man, are you a genius?

nrlwinner · Jan 27, 2010

Got another one that I can't do.

Prove

$|1-zW|^2-|z-w|^2=(1-|z|^2)(1-|w|^2)$
for any pair of complex numbers z and w, where W is the conjugate of w

untouchablecuz · Jan 28, 2010

untouchablecuz · Jan 28, 2010

nrlwinner said:
Man, are you a genius?

=$

nrlwinner · Jan 28, 2010

I'm stuck on another question

$$The points A, B and C form a triangle whose vertices present the complex numbers \alpha ,\beta and \gamma respectively.$

I've proved that if
$\frac{\gamma -\beta }{\gamma -\alpha }=Cis\frac{\pi }{3}$ then it's an equilateral triangle.

But I need help to show

$\alpha ^2+\beta ^2+\gamma ^2=\beta \gamma +\gamma \alpha +\alpha \beta$

untouchablecuz · Jan 28, 2010

nrlwinner said:
I'm stuck on another question

$$The points A, B and C form a triangle whose vertices present the complex numbers \alpha ,\beta and \gamma respectively.$

I've proved that if
$\frac{\gamma -\beta }{\gamma -\alpha }=Cis\frac{\pi }{3}$ then it's an equilateral triangle.

But I need help to show

$\alpha ^2+\beta ^2+\gamma ^2=\beta \gamma +\gamma \alpha +\alpha \beta$

$\\\frac{\gamma -\beta }{\gamma -\alpha }=cis(\frac{\pi}{3}) \ (*) $ and $\frac{\gamma -\alpha }{\gamma -\beta }=cis(-\frac{\pi}{3}) \ (**)\\(*)+(**),\\\frac{\gamma -\beta }{\gamma -\alpha }+\frac{\gamma -\alpha }{\gamma -\beta }=cis(\frac{\pi}{3})+cis(-\frac{\pi}{3})=2\Re \left(cis(-\frac{\pi}{3}) \right )=1\\$Multiply through by $(\gamma -\alpha)(\gamma - \beta),\\(\gamma -\beta)^2+(\gamma -\alpha)^2=(\gamma -\alpha )(\gamma -\beta )\\\alpha ^2+\beta ^2+\gamma ^2=\alpha \beta +\alpha \gamma +\gamma \beta$

gurmies · Jan 28, 2010

nrlwinner said:
I'm stuck on another question

$$The points A, B and C form a triangle whose vertices present the complex numbers \alpha ,\beta and \gamma respectively.$

I've proved that if
$\frac{\gamma -\beta }{\gamma -\alpha }=Cis\frac{\pi }{3}$ then it's an equilateral triangle.

But I need help to show

$\alpha ^2+\beta ^2+\gamma ^2=\beta \gamma +\gamma \alpha +\alpha \beta$

You've posted this question already...although this disguises it quite nicely.

nrlwinner · Jan 28, 2010

$$If the points z_1, z_2, z_3 lie\ on\ a\ circle\ through\ the\ origin,\ prove\ that\ the\ points\ \frac{1}{z_1},\frac{1}{z_2},\frac{1}{z_3}\ are\ collinear$

Can someone start me off, because I've got no clue what to do, but not finish it. I wanna have a go.

Also need help with the last question of my paper, and then I'm done!!! I've tried employing the strategies you guys showed me in the previous questions, but this one puts me off because there are 3 brackets, and the vectors are reversed.

z1,z2,z3 are represented by the points p1,p2,p3 respectively. If p1p2p3 is an equilateral triangle, show

$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$

nrlwinner · Jan 28, 2010

bobt2062 said:
I worked out a way to do your first question, but I don't know how to describe it in part. The only clue I can give is that at starts out as a circle geometric proof, and by playing with Arguments, turns algebraic. Tell me if you want to see it.

Yeah, I was on that track but didn't think it was right. May I see the solution?

nrlwinner · Jan 29, 2010

bobt2062 said:
Well, I don't know how to draw a diagram here, so I'll have to describe it for you.

Draw your circle through O, predominantly in the first quadrant. Let A, B, C represent z1, z2, z3 respectively. For ease of describing, draw O, A, B, C in clockwise order (and in left to right order), all in the first quadrant. (The same logic works if they are drawn otherwise.) Produce AB to D.

Angle AOC = arg z1 - arg z3.

Angle CBD = arg (z1 - z2) - arg (z2 - z3)
(Tell me if you don't get that)

So arg (z1 - z2) - arg (z2 - z3) = arg z1 - arg z2 [ext angle of cyclic quad = opp int angle]

Rearrange:
arg(z1 - z2) - arg z1 = arg (z2 - z3) - arg z3

Add in the same 'fudge factor' on both sides of the =:
arg (z1 - z2) - arg z1 - arg z2 = arg (z2 - z3) - arg z3 - arg z2

$\arg \left (\frac{z_{1}-z_{2}}{z_{1}z_{2}}\right )= \arg \left (\frac{z_{2}-z_{3}}{z_{2}z_{3}}\right )$

$\arg \left (\frac{1}{z_2}- \frac{1}{z_1}\right)= \arg \left (\frac{1}{z_3}- \frac{1}{z_2}\right)$

Can you see that it is now proved?

What's D?

gurmies · Jan 29, 2010

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