Author Topic: Specialist Problem (Read 11062 times) Tweet Share

another problem..

(a) If v = 1 + i and z = x + iy, for any real numbers x and y:
(i) Show that the equation
|z − v| = |vz|
represents a circle, and find its centre and radius.
(ii) Find the intersection of the circle in part (i) with the straight line
|z − v| = |z + v| .

(b) Using the roots of z^5 = 1, or otherwise, write z^4 + z^3 + z^2 + z + 1 as the product of two quadratic
expressions with real coefficients.
Hence find the exact value of the product $cos\frac{2\pi}{5}cos\frac{4\pi}{5}$

ed_saifa · « **Reply #38 on:** April 05, 2008, 09:42:12 am »

$|z-v|=|vz|$
Let z=x+yi
$|x+yi-(1+i)|=|(1+i)(x+yi)|$
Group real and imaginary parts
$|x-1+yi-i|=|x-y+yi+ix|$
$|(x-1)+i(y-1)|=|(x-y)+i(x+y)|$
Find the moduli
$\sqrt{(x-1)^2 +(y-1)^2}=\sqrt{(x-y)^2 + (x+y)^2}$
Square both sides
$(x-1)^2 + (y-1)^2=(x-y)^2 + (x+y)^2$
Expand and collect up terms
$x^2-2x+1+y^2-2y+1=x^2-2xy+y^2+x^2+2xy+y^2$
$x^2+y^2+2x+2y-2=0$
Factorise
$(x^2+2x+1)-1+(y^2+2y+1)-1-2=0$
$(x+1)^2 + (y+1)^2= 4$
centre: (-1,-1) r=2

khalid!!! · « **Reply #39 on:** April 05, 2008, 09:57:25 am »

Quote from: Ahmad on February 24, 2008, 08:14:20 am

$(\cos \theta + i \sin \theta)^4 = \cos 4\theta + i \sin 4\theta$

Expand the left hand side and equate the real and imaginary parts of the expanded left hand side and the right hand side.

-_- i dont get it xD

ice_blockie · « **Reply #40 on:** April 05, 2008, 10:16:38 am »

I want your stationery Neobeo!!!! Where did you get it?

Mao · « **Reply #41 on:** April 05, 2008, 12:14:18 pm »

ice blockie:

Quote from: ice_blockie on April 05, 2008, 10:16:38 am

I want your stationery Neobeo!!!! Where did you get it?

he uses a tablet PC xD

ed_saifa:

Quote from: ed_saifa on April 05, 2008, 09:42:12 am

$\sqrt{(x-1)^2 +(y+1)^2}=\sqrt{(x-y)^2 + (x+y)^2}$

that should be a $y-1$

Tea.Bag:
a) ii)

$|z-v|=|vz|$ : $(x+1)^2 + (y+1)^2= 4$

$|z-v|=|z+v|$

$\implies \sqrt{(x-1)^2+(y-1)^2}=\sqrt{(x+1)^2+(y+1)^2$

$\implies x^2-2x+1+y^2-2y+1=x^2+2x+1+y^2+2y+1$

$\implies -4x-4y=0$

$\implies y=-x$

Hence the intersection between $|z-v|=|vz|$ and $|z-v|=|z+v|$ is when both relationships are satisfied:
substituting $y=-x$ to our circle:

$(x+1)^2 + (-x+1)^2= 4$

$x^2+2x+1+x^2-2x+1=4$

$2x^2=2$

$x^2=1$

$x=\pm 1$

$y=\mp 1$

$z=\pm1 \mp i$

b)
it should be noted that:
$x^n-a^n=(x-a)\left(x^{n-1}+a\cdot x^{n-2}+a^2 \cdot x^{n-3}+ \cdots + a^{n-2}\cdot x + a^{n-1}\right)$

The roots of $z^5=1$ :

$|z|^5cis(5\theta)=cis(0+2k\pi )$

$\implies |z|=1$

$\implies \theta = 0,\frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \frac{8\pi}{5}$

$\implies Arg(z)=-\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}$

$z=1,cis\left(\pm \frac{2\pi}{5}\right), cis\left(\pm \frac{4\pi}{5}\right)$

This means the expression $z^5-1$ can be expressed as:

$z^5-1=(z-1)\left(z-cis\left(\frac{2\pi}{5}\right)\right)\left(z-cis\left(-\frac{2\pi}{5}\right)\right)\left(z-cis\left(\frac{4\pi}{5}\right)\right)\left(z-cis\left(-\frac{4\pi}{5}\right)\right)$

since the roots above are all conjugates of each other:

$z^5-1=(z-1)\left(z^2-\left(cis\left(\frac{2\pi}{5}\right)+cis\left(-\frac{2\pi}{5}\right)\right)z+|z|\right)\left(z^2-\left(cis\left(\frac{4\pi}{5}\right)+cis\left(-\frac{4\pi}{5}\right)\right)z+|z|\right)$

$z^5-1=(z-1)\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

$(z-1)\left(z^4+z^3+z^2+z+1\right)=(z-1)\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

$z^4+z^3+z^2+z+1=\left(z^2-2\cos\left(\frac{2\pi}{5}\right)z+1\right)\left(z^2-2\cos\left(\frac{4\pi}{5}\right)z+1\right)$

Hence: expanding the RHS:

$\implies z^4-\left(2\cos\left(\frac{4\pi}{5}\right)+2\cos\left(\frac{2\pi}{5}\right)\right)z^3+\left(2+4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)\right)z^2-\left(2\cos\left(\frac{4\pi}{5}\right)+2\cos\left(\frac{2\pi}{5}\right)\right)z+1$

therefore, by equality of polynomials:

$z^2=\left(2+4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)\right)z^2$

$4\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)=-1$

$\cos\left(\frac{2\pi}{5}\right)\cos\left(\frac{4\pi}{5}\right)=-\frac{1}{4}$

chid · « **Reply #42 on:** April 05, 2008, 01:27:39 pm »

I've got a question.

Arg(z)= π/4 and Arg(z-3)= π/2. Find Arg(z-6i)

I'm a little confused.

Thanks.

ed_saifa · « **Reply #43 on:** April 05, 2008, 02:18:39 pm »

Thanks Mao I corrected it XD

Tea.bag · « **Reply #44 on:** April 09, 2008, 05:37:15 pm »

another problem...

Find the range of $f(x)=cos(sin^{-1}2x)$

Search the forums now!

We have moved!

Author Topic: Specialist Problem (Read 11062 times) Tweet Share

Neobeo

Re: Specialist Problem

ed_saifa

Re: Specialist Problem

Ahmad

Re: Specialist Problem

Mao

Re: Specialist Problem

Collin Li

Re: Specialist Problem

Neobeo

Re: Specialist Problem

Tea.bag

Re: Specialist Problem

Tea.bag

Re: Specialist Problem

ed_saifa

Re: Specialist Problem

khalid!!!

Re: Specialist Problem

ice_blockie

Re: Specialist Problem

Mao

Re: Specialist Problem

chid

Re: Specialist Problem

ed_saifa

Re: Specialist Problem

Tea.bag

Re: Specialist Problem

Recent Posts

User:
Password: