Complex Number Questions (2 Viewers)

[qqmore]

Hello,

I'm having trouble doing 3c)ii). I showed that y could be y = 0 or x^2+y^2 = 1, but I have no clue on how to prove |k|> or < 2. Likewise, for qns 3b).... how do u actually explain |z+2| or arg (z+2) is between those boundaries?


Any help will be apprecipated,

thx.

 

[undalay]

just draw the diagram : O?

 

[qqmore]

thanks alot for 3b)ii) ..... how do u show for 1 \< |z+2| \< 3 though?

 

[undalay]

1 \< |z+2| \< 3


the bigger circle is that graph.

edit: biggest.

 

[qqmore]

thanks a lot, i understand 3b) now... but i still need help on 3c) =(

 

[namburger]

i)
z + 1/z = (z^2+1)/z
= (x^2 + 2ixy + y^2 + 1)/(x+iy)
= (x^2 + 2ixy + y^2 + 1)(x-iy)/(x^2+y^2)
= [(x^3+xy^2+x) + (yx^2+y^3-y)i]/(x^2+y^2)

if k is real, than fake = 0
(yx^2+y^3-y)/(x^2+y^2) = 0
(yx^2+y^3-y) = 0
y(x^2+y^2-1) = 0
Therefore y=0 or x^2+y^2 = 1

ii)
(x^3+xy^2+x)/(x^2+y^2) = k
If y = 0, then:
(x^3 + x)/x^2 = k
x^2+1 = kx
x^2 - kx + 1 = 0
b/c k is real, delta > 0
K^2 >/ 4
|K| >/ 2

If x^2+y^2 = 1 then:
x(x^2+y^2+1) = k
2x = k

By inspection, k/<2 because from (x^2+y^2 = 1), x has to be less than 1

 

[conics2008]

i)
z + 1/z = (z^2+1)/z
= (x^2 + 2ixy + y^2 + 1)/(x+iy)
= (x^2 + 2ixy + y^2 + 1)(x-iy)/(x^2+y^2)
= [(x^3+xy^2+x) + (yx^2+y^3-y)i]/(x^2+y^2)

if k is real, than fake = 0
(yx^2+y^3-y)/(x^2+y^2) = 0
(yx^2+y^3-y) = 0
y(x^2+y^2-1) = 0
Therefore y=0 or x^2+y^2 = 1

ii)
(x^3+xy^2+x)/(x^2+y^2) = k
If y = 0, then:
(x^3 + x)/x^2 = k
x^2+1 = kx
x^2 - kx + 1 = 0
b/c k is real, delta > 0
K^2 >/ 4
|K| >/ 2

If x^2+y^2 = 1 then:
x(x^2+y^2+1) = k
2x = k

By inspection, k/<2 because from (x^2+y^2 = 1), x has to be less than 1

Dam you ! angry face.. you were faster.. i had to write it on the paper first then post it up, how fast do you do it on the net damm u LoL.. the question say if k is real then fake = 0

nice one buddy, ill try better next time =)

the 2nd quesiton seem straight forward just skectch two circles one with radisu 1 and the other with root of 3/

and the other region basically is a straight line..

 

[qqmore]

cheers, never thought to equate k and x to form an equation><

 

[namburger]

i meant imaginary = 0,
haha my friend called it fake once, think i picked it up from him

 

[conics2008]

i meant imaginary = 0,
haha my friend called it fake once, think i picked it up from him

i call it fake too, it makes life much more easier =)

 

[qqmore]

Thank you for all your help for the previous question,

I just stumbled upon another question I'm not sure how to do:
I managed to to i) but not sure how to do ii)....
Any hints/clues/working outs are welcomed

 

[ronnknee]

Similarly, equate the imaginary parts

 

[namburger]

hey buddy, what program do you use to write it up. Getting tired of typing it up

Fake and imaginary mean the same thing ronnkneeeeeee =]

 

[conics2008]

hey guys how does cos(x) = 2cos^2 x/2 -1 ???

what formula is this ?? never came across this

 

[pLuvia]

It is cos2x=2cos²x-1

But replacing the x in the 2x with an x/2

 

[ronnknee]

Hahaha "fake", I laughed when I read it in your working out
Sif make up your own terms =p

Ahh I write my solutions in Photoshop with a graphics tablet, then I use Paint Shop Pro to crop and decrease the colour depth

And yea, that's the double/half angle formula

 

[conics2008]

its not in the sylb ?????????

 

[Forbidden.]

Why the hell does the Board Of Studies use the 'cis' notation and what is it ?

 

[tacogym27101990]

yes its in the syllabus
its just like
cos2x = cos(x+x) = cos²x - sin²
then by rearannging it u get 2cos²x - 1
but his just used cos(x/2+x/2)
surely u know this conics?

 

[ronnknee]

cos 2x = 2cos^2 x - 1
Let x = u/2
cos u = 2cos^2 u/2 - 1