questions for the smart ppl (mainly complex nos some graphs and conics) (1 Viewer)

[.ben]

here it is sorry about that.

 

[pLuvia]

Which ones still need to be done ben?

 

[.ben]

questions 1 and 2 for conics, questions 3, 4,7, and 9 for complex numbers.

thanks again

 

[Riviet]

CORRECTION TO QUESTION 9!!!

9.

let w=x+iy and z=X+iY

Since |w|=1, then x²+y²=1

So z=(x+iy+1)/[(1-x)-iy]

= {[(x+1)+iy][(1-x)+iy]} / (1-2x+x²+y²)

= (x-x²+ixy+1-x+iy+iy-ixy-y²) / (2-2x)

= 2iy/(2-2x)

= iy/(1-x)

equating real and imaginary parts of w and z,

Y²/(1-X)² = 1

Y=+sqrt[(1-X)²]

Y=+|1-X|

.'. the locus of z are the absolute value functions Y=|1-X| and Y=-|1-X|

I hope that's right.

 

[Trev]

3.
arg[(z-i)/(z+i)]=pi/4
This defines the locus point, it is a right angle from both points (0,i) and (0,-i) on the argand diagram.
The locus is therefore a circle with equation x²+y²=1.
(by the definition that the angle subtended from a semi-circle to the circumference is a right angle).

 

[.ben]

ok thanks any more?

 

[Riviet]

I have no idea of where to go in 7, and I can't seem to prove the first bit in 4, can't get the 2cos(theta/2).

 

[Mountain.Dew]

okay, heres my 2 cents:

for (1)

its a simple case of getting the chord of contact from (3,4) --> which happens to be 4y/9 + 3x/16 = 1
I THINK

then, simulatenous equations with the ellipse --> find the 2 pts the chord intersects (say at P and Q), call (3,4) pt R --> then its a simple cas of PR * PQ = -1.

(4) --> r u sure ur trying to get that result with the 1+.... inside? it does look impossible...

(7) --> simply get a common denominator --> do a little bit of expanding and simplifying in the numerator, using the fact that 1 + w + w^2 = 0 and w^3 = 1. must also realise that the denominator, (z-1)(z-w)(z-w^2) = z^3 - 1, since 1, w and w^2 are the ROOTS of z^3 - 1.

if i can get some more, ill post em up

 

[Mountain.Dew]

AAAAAAAAAAAAAAHHHHH revelation!

for (2) --->

eventually u will get something like:

dy/dx = (3x^2 + 4xy + 3y^2) / (2x^2 + 6xy + 3y^2)

so, u wanna get dy/dx = y/x SO, do this:

dy/dx = (3x^2 + 4xy + 3y^2) / (2x^2 + 6xy + 3y^2) * (x/y) * (y/x) --> expand the x/y part wth the large algebraic term, and substitute x^2y = (x+y)^3 --> do some expanding and simplifying, and u will eventually get dy/dx = y/x.

this seems to be a tried and true method, but there SHOULD be a quicker way...

 

[Riviet]

Ah, nice observations for 7 M.D, I think I've got it now:

Observing that 1, w and w² are the roots of z³-1,

ie (z-1)(z-w)(z-w²)=z³-1

and (z-w)(z-w²)=(z³-1)/(z-1)=z²+z+1

LHS= [(z-w)(z-w²)+w(z-1)(z-w²)+w²(z-1)(z-w)] / (z-1)(z-w)(z-w²)

= (z²+z+1+wz²-z-wz+1+w²z²-z-wz²+1) / (z³-1)

=[z²(1+w+w²) + 3 -z(1+x+w²)] / (z³-1)

=3/(z³-1) ~ QED

 

[wogboy]

An alternative way to do 1b):

Let O = the origin, C = circle's centre, T = point where tangent meets circle.

angle OTC = 90 deg (tangent of a circle is perpendicular to the radius)
CT = 1
OC = 2

-> angle COT = arcsin(1/2) = 30 deg

so the minimum argument of z is 60 deg = pi/3 radians