4U Revising Game (1 Viewer)

[namburger]

Show that the roots of z⁶ + z³ + 1 = 0 are among the roots of z⁹ - 1 = 0. Hence find the roots of z⁶ + z³ + 1 = 0 in modulus/argument form

z^9 -1 = (z^3-1)(z^6 + z^3 + 1) {Using different of cubes}

z^9 = 1
cis 9@ = cis 2npi
@ = 2npi/9
@ = -/+ [2pi/9, 4pi/9, 6pi/9, 8pi/9 and 0]
Since cis +/-(6pi/9) and cis 0 are roots of (z^3-1), therefore cis of +/- [2pi/9, 4pi/9 8pi/9] are the roots

My previous q:
Complex no.: Sketch the locus of the point z:
arg [(z-5)/(z+1)] = pi/4. Find its cartesian equation

 

[ronnknee]

My previous q:
Complex no.: Sketch the locus of the point z:
arg [(z-5)/(z+1)] = pi/4. Find its cartesian equation

 

[ronnknee]

Hahha I forgot how to do it T_T
I spent ten minutes trying and I couldn't
Someone can help me find it xD

 

[ronnknee]

Yea I was thinking of the theorem "angle at centre twice the angle at circumference". That means the angle subtended by the centre and the two points is a right angle; thus it's a right angle triangle. The chord length is 6

36 = 2r[sup2[/sup]
18 = r[sup2[/sup]
r = 3 root 2

Let the equation of the circle be (x - h)² + (y - k)² = r²
Since (-1, 0) and (5, 0) satisfy the equation,
(x + 1)² = (x - 5)²
2x + 1 = -10x +25
x = 2
Therefore h = 2


(x - 2)² + (y - k)² = 18

That's as much as my head allows me to do at 12:45 in the morning :|

 

[namburger]

rofl you make it tooo complex ronknee.

Hope this helps:
http://i237.photobucket.com/albums/ff3/Namburger/Complexnoq.jpg

 

[ronnknee]

Hahaha I honestly forgot how to do that question
It's not like I'm complex2008 D:
But yea, that diagram sure helped, thanks =p

(x + 2)² + (y - 3)² = 18

 

[namburger]

must put y>0, i lost a mark because of that =[

 

[ronnknee]

Question 10 (Volume)
A solid figure has as its base, in the xy plane, the ellipse x2/16 + y2/4 = 1

Cross sections perpendicular to the x-axis are parabolas with latus rectums in the xy plane. Find the volume of the solid

x⁴ + px³ + qx² + rx + s = 0 has two roots which are reciprocals and two other roots which are opposites. Show that q = 1 + s and r = ps

 

[Poad]

x⁴ + px³ + qx² + rx + s = 0 has two roots which are reciprocals and two other roots which are opposites. Show that q = 1 + s and r = ps

Let roots be: a, 1/a, b, -b.

Product of roots
a.(1/a).b.-b = e/a
-b² = s
Sum of roots 2 at a time
(a.1/a) + ab - ab + (b/a) - (b/a) - b.b = c/a
1 - b² = q
.:. 1 + s = q

And, uh, someone else can do the other bit. >_>

 

[ronnknee]

Hahhaha it's volumes =p

 

[Trebla]

Find ∫ x⁹e^x5 dx

 

[namburger]

Find ∫ x⁹e^x5 dx

Let u = x^5
du = 5x^4 dx

∫ x⁹e^x⁵ dx
= 1/5 ∫ue^u du

Do it by parts

 

[Yip]

∫ x⁹e^x⁵ dx
= 1/5 ∫ue^u du

Do it by parts

There is also another curious way to do this integral, I'll leave another peson to do the by parts method, here is the other method:

Consider f(t)=∫e^(ut)dt
df/dt=∫ue^(ut)dt

Also, f(t)=[e^(ut)]/u so df/dt=[ute^(ut)-e^(ut)]/t^2
Therefore ∫ue^(ut)dt=[ute^(ut)-e^(ut)]/t^2
Letting t=1 gives ∫ue^u du=ue^u-e^u

∫ x⁹e^x⁵ dx=[1/5][(x^5)e^(x^5)-e^(x^5)]

Here is a interesting integral:

Calculate ∫[ln(1+x)]/(1+x^2)dx over the interval 0< x <1

 

[namburger]

Here is a interesting integral:

Calculate ∫ln(1+x)/(1+x^2)dx over the interval 0< x <1

By parts again?
∫ln(1+x)/(1+x^2) dx

u = ln(1+x)
du = dx/(1+x)

dv = 1/(1+x^2)dx
v = arctan x

∫ln(1+x)/(1+x^2) = [ln(1+x) . arctan x] - ∫arctan/(1+x) dx
= ln 2 . pi/4 - ∫arctan/(1+x) dx

Apparently you can't integrate that function =[
Unless you meant ln(1+x) - ln(1+x^2)?

 

[Yip]

No, my question is typed correctly. That integral has no closed form solution if its indefinite, however, it has a nice exact answer if u integrate it over 0 to 1. Integration by parts is going to get u nowhere, u have to try a different approach.

 

[PF]

By parts again?
∫ln(1+x)/(1+x^2) dx

u = ln(1+x)
du = dx/(1+x)

dv = 1/(1+x^2)dx
v = arctan x

∫ln(1+x)/(1+x^2) = [ln(1+x) . arctan x] - ∫arctan/(1+x) dx
= ln 2 . pi/4 - ∫arctan/(1+x) dx

Apparently you can't integrate that function =[
Unless you meant ln(1+x) - ln(1+x^2)?

hint: never do the suggested "interesting" questions, especially when it comes from yip

 

[namburger]

hint: never do the suggested "interesting" questions, especially when it comes from yip

Gotta keep trying =]

∫ln(1+x)/(1+x^2) dx

Let x = tan @
dx = sec^2 @ d@

When x = 1, @=pi/4
x = 0, @ = 0

Therefore:
∫ln(1+ tan @) d@ between 0 --> pi/4
using the fact that ∫f(x) = ∫f(a-x)

∫ln(1+ tan @) d@ = ∫ln(1+ tan (pi/4 - @)) d@
= ∫ln(1+ (1-tan@)/(1+tan@)) d@
= ∫ln(2/(1+tan@))
= ∫ln 2 - ∫ln(1+tan@)

Therefore:
2 ∫ln(1+tan@) = ∫ln 2
∫ln(1+tan@) = ln 2 . pi/8

 

[Poad]

Let roots be: a, 1/a, b, -b.

Product of roots
a.(1/a).b.-b = e/a
-b² = s
Sum of roots 2 at a time
(a.1/a) + ab - ab + (b/a) - (b/a) - b.b = c/a
1 - b² = q
.:. 1 + s = q

And, uh, someone else can do the other bit. >_>

Oops, just realised how silly it was of me to use 'a' to represent both the coefficient of x⁴ and the root. Silly me.

 

[conics2008]

Heres a complex numbers question.

If | z- 6i| = 5

Prove that arg ( z-4-3i / z+3-2i ) = pi/4 or 3pi/4

 

[conics2008]

locus of z is a circle radius 5 centered at (0,6)
equation of locus: x^2 + (y-6)^2 = 5^2
both (4,3) and (-3,2) satisfy the equation, and lie on the locus.

arg ( z-4-3i / z+3-2i ) is therefore the angle made between the two points (4, 3) and (-3, 2) standing on an arc of a circle to its circumference.

distance between these points = sqrt [ (4 - - 3)^2 + (3 - 2)^2] = sqrt (50)
since sqrt [(radius)^2 + (radius)^2] = sqrt (5^2 + 5^2) = sqrt (50)
pythagoras implies that the angle made between the points and the center is pi/2

since angle on circumference is double the angle made at center
arg ( z-4-3i / z+3-2i ) = pi/4 (angles above the line which goes through points)

or arg ( z-4-3i / z+3-2i ) = 3pi/4 (angles below the line which goes through points as 3pi/4 = pi - pi/4 from cyclic quad)

the other thingy doesn't have to be standing on the circumference, show me your diagram..

you can also have the circle and those two lines which make pi/4 or 3pi/4 be made at the center of the circle.. sorry, i dont know how to draw good on paint, it turns out really bad. post your diagram.