Complex (1 Viewer)

[The Nomad]

Can someone start me off, because I've got no clue what to do, but not finish it. I wanna have a go.

Also need help with the last question of my paper, and then I'm done!!! I've tried employing the strategies you guys showed me in the previous questions, but this one puts me off because there are 3 brackets, and the vectors are reversed.

z1,z2,z3 are represented by the points p1,p2,p3 respectively. If p1p2p3 is an equilateral triangle, show

Refer to the last question of the Terry Lee vectors exercise for a purely algebraic answer.

 

[nrlwinner]

Well, I don't know how to draw a diagram here, so I'll have to describe it for you.

Draw your circle through O, predominantly in the first quadrant. Let A, B, C represent z1, z2, z3 respectively. For ease of describing, draw O, A, B, C in clockwise order (and in left to right order), all in the first quadrant. (The same logic works if they are drawn otherwise.) Produce AB to D.

Angle AOC = arg z1 - arg z3.

Angle CBD = arg (z1 - z2) - arg (z2 - z3)
(Tell me if you don't get that)

So arg (z1 - z2) - arg (z2 - z3) = arg z1 - arg z2 [ext angle of cyclic quad = opp int angle]

Rearrange:
arg(z1 - z2) - arg z1 = arg (z2 - z3) - arg z3

Add in the same 'fudge factor' on both sides of the =:
arg (z1 - z2) - arg z1 - arg z2 = arg (z2 - z3) - arg z3 - arg z2





Can you see that it is now proved?

I can't seem to se how the last line proves they're collinear.

 

[Trebla]

If you think about it in vector terms, if the arguments are equal they are at least parallel. However, since the complex number 1/z₂ is common in both equalities then the vectors must have a common point as well as be parallel which implies they must be collinear.

 

[nrlwinner]

Thanks.I got it now.

Also.

For the polynomial z^4 -2z^3 +7z^2 -4z +10,
find real values of a for which ai is a solution.

I know it has something to do with conjugate root theorem, but I'm not sure how to apply it without actual numbers.

 

[shaon0]

Thanks.I got it now.

Also.

For the polynomial z^4 -2z^3 +7z^2 -4z +10,
find real values of a for which ai is a solution.

I know it has something to do with conjugate root theorem, but I'm not sure how to apply it without actual numbers.

Let P(x)=z^4-2z^3+7z^2-4z+10
P(ai)=a^4+2a^3.i+7a^2-4ai+10
P(-ai)=a^4-2a^3+7a^2+4ai+10
P(ai)+P(-ai)=2(a^4-7a^2+10)=0
(a^2-5)(a^2-2)=0
a=+-sqrt(5) and +-sqrt(2)
Let A,B,C,D be roots of P(z):
A+B+C+D=2
C+D=2 and CD=5 or 2
C^2+D^2=(C+D)^2-2CD
ie. C^2+D^2=-6 or 0 but we know C=/=+-iD as this would not satisfy poly equ
Thus, CD=5
C^2-2C+5=0
(C-1)^2=-4
C=1+2i, D=1-2i

Thus, z=+-sqrt(2)i or 1+-2i
ie a=+-sqrt(2)

 

[untouchablecuz]

alternatively

 

[shaon0]

alternatively

Nice, a lot better than my solution but i think i could've just cut-off a=+-sqrt(5) before hand as a solution

 

[nrlwinner]

I need help on this question.

I did the first part and got |z|=2 as a locus

But what is the locus of w where

 

[cutemouse]

I need help on this question.

I did the first part and got |z|=2 as a locus

But what is the locus of w where

 

[khorne]

Algebraically:

a^2 + b^2 = 4

w = a+ib - 1 / a+ib, after simplifying:

w = 4-a+ib/4

w = x+iy, so equate:

x = 4-a/4 and y = b/4

a^2 = 16(x-1)^2
b^2 = 16y^2

resub:

16(x-1)^2 + 16y^2 = 4

(x-1)^2 + y^2 = 1/4

|w-1| = 1/2

 

[nrlwinner]

Thanks. I need help finishing off this question.

I've found the five-fifth roots of uity and proved that if w=cis 2pi/5
1+w+w^2+w^3+w^4=0

But I cant show that w+w^4 and w^2+w^3 are the roots of z^2+z-1

 

[ninetypercent]

if w + w^4 and w^2 + w^3 are roots of the equation, then

sum of roots = w + w^4 + w^2 + w^3 = -1
product of roots = (w + w^4)(w^2 + w^3) = w^3 + w^6 + w^4 + w^7 = w^3 + w^4 + w + w^2 = -1

the equation is
z^2 - (sum of roots) z + product of roots = 0
z^2 + z -1 = 0

which is the equation given in question

 

[khorne]

What are you doing up at 11pm doing 4U?

 

[ninetypercent]

^ he might have tutor hw or something due.

 

[nrlwinner]

I know this sound so simple, but I've never had to find the roots (apart for square) for something in cartesian form. I've always done it either just as a real number of just as an imaginary number.

Find the five-fifth roots of

and I'm tryna finish all my complex work tonight and I've got like 5 questions out of 153 I can't do

 

[ninetypercent]

the trick is to convert sqrt 3 + i into mod-arg form

 

[nrlwinner]

Thank you. And final question before I'm finished all this sh*t.

 

[khorne]

is that 1cos2@ or 1+cos2@

 

[nrlwinner]

sorry its 1+cos2@

 

[ninetypercent]

btw you typed the question out incorrectly, but I was able to deduce what the question was by looking at the required result

By Demoivre's Theorem