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questions for the smart ppl (mainly complex nos some graphs and conics) (1 Viewer)
- Thread starter .ben
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pLuvia
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Which ones still need to be done ben?
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.
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.
Last edited:
Trev
stix
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 x<sup>2</sup>+y<sup>2</sup>=1.
(by the definition that the angle subtended from a semi-circle to the circumference is a right angle).
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 x<sup>2</sup>+y<sup>2</sup>=1.
(by the definition that the angle subtended from a semi-circle to the circumference is a right angle).
Mountain.Dew
Magician, and Lawyer.
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
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
Magician, and Lawyer.
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...
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...
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
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
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
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