complex number question (1 Viewer)

[pLuvia]

Solve
z³ - 5z² + 8z - 6 = 0, a root is 1 - i

 

[香港!]

Solve
z³ - 5z² + 8z - 6 = 0, a root is 1 - i

z=3, 1+i, 1-i

 

[pLuvia]

Can you show me how you got that please?

 

[香港!]

well the equation is a polynomial w\ real coefficients
so by conjugate root theorem, 1-i is a root so 1+i is also a root
then by inspection z=3
so u got the three roots

 

[pLuvia]

So you just use the conjugate of one of the roots and use factor theorem?

 

[Trev]

Or; since 1+i and 1-i are both roots:
Divide z³-5z²+8z-6=0 by (z-(1+i))(z-(1-i)) since it is a factor.

 

[香港!]

So you just use the conjugate of one of the roots and use factor theorem?

u can do that..
but u can also do dis:
sum of roots
let A be the other root
(1-i)+(1+i)+A=5
2+A=5
A=3

 

[pLuvia]

This is the example solution it says

z = 1 - i is a root, z = 1 + i is another root.

They correspond to the factor z² - 2z + 2.

I don't understand that

 

[香港!]

^ That's what Trev did...

 

[pLuvia]

Ah, nevermind I understand it now, thanks guys

 

[pLuvia]

(a) If x and y are real numbers such that
(x+yi)² = 4+3i
(i) find real numbers a and b such that
(x-yi)² = a+bi

(b) Solve the equation
(z²-4)² + 9 = 0

(a)
(i) a = 4 and b = -3
(ii) x = ±3/√2 and y = ±1/√2

(b) Can't do

ANSWERS:
(a)
(i) a = 4 and b = -3
(ii) x = ±3/√2 and y = ±1/√2

(b) ±3/√2 ±(1/√2)i

 

[yrtherenonames]

Part b) is a complex difference of two squares, just like you can factorise x^2+1=(x+i)(x-1), so you can do the same thing with this.
So (x^2-4+3i)(x^2-4-3i)=(x^2-4)^2+9
Then u use part (i) to find the square roots of 4+3i/4-3i and ure home shweet

 

[Trev]

(x+yi)²=4+3i
x²+y²+2xyi=4+3i
Taking real and unreal parts:
x²+y²=4
2xy=3; solve simultaneously.
Same method for part (i)

b)
(z²-4)²=-9
z²-4=+/-3i
z²=4+/-3i; let z=a+bi
(a+bi)²=4+3i, and (a+bi)²=4-3i; expand and solve simultaneously like in (a); a different method to what is posted above.

 

[Riviet]

Ah i'm bored here, so i'll just finish off trev's solution:
(a+bi)²=4+3i
a²-b²+2abi=4+3i
equate real and imaginary parts to get 2 equations
a²-b²=4 (1)
ab=3/2 (2)
from (2),
b=3/2a (3)
sub (3) in (1)
a²-(3/2a)²=4
a²-9/4a²=4
4a⁴-16a²-9=0
(2a²-9)(2a²+1)=0
a=+/- 3/sqrt2 (rationalising denom. gives +/- 3sqrt2/2)
sub a values into (3)
b=+/- sqrt2/2
.: sqrt(4+3i)= [3sqrt2+(sqrt2)i]/2, [-3sqrt2-(sqrt2)i/]2
sqrt(4-3i) would be very similar, too so just follow the same procedure.
(I'm not typing all this again... too tired -_-)

 

[justchillin]

The easiest way to do this question is to recognise that the complex conjugate is a root..then find a quadratic factor then find the other factor by inspection (ie see that u have to put to get the 1st and last terms of the polynomial...

 

[pLuvia]

Deduce that if @ is a non-real root of ax² + bx + c = 0, where a,b,c are real, then @-bar is the other root of this quadratic equation

 

[Trev]

Complex conjugate pairs... (Do you need a proof? Because if you do I don't have one )

 

[Stefano]

Deduce that if @ is a non-real root of ax² + bx + c = 0, where a,b,c are real, then @-bar is the other root of this quadratic equation

@ = {-b +/- root.[b²-4ac]}/2a (Quadratic Formula)

Therefore, the roots are:

1. {-b + root.[b²-4ac]}/2a (@)

2. {-b - root.[b²-4ac]}/2a (@-bar)


Understood?

 

[KFunk]

This doctor math thing has an interesting bit of reasoning for it: http://mathforum.org/library/drmath/view/53874.html

EDIT: Maybe I should have read my text book , it seems that the Arnold one has a decent little proof which, abbreviated, goes like this:

Let P(x) be a polynomial with real coefficients and a complex zero α where α = a +bi and αbar = a - bi.

Then consider D(x) = (x - α )(x - αbar) = x² - 2Re(α )x + |α|²

Using the division transformation you can express P(x) as P(x) = D(x).Q(x) + R(x) and deg R < deg D = 2.

P(x) = (x - &alpha; )(x - &alpha;bar)Q(x) + cx + d

P(&alpha; ) = c&alpha; + d , and P(&alpha;bar) = c(&alpha;bar) + d = (c&alpha; + d)bar = [P(&alpha; )]bar

Hence it follows that if P(&alpha; ) = 0 then P(&alpha;bar)=0

 

[acmilan]

You can prove that complex roots come in conjugate pairs:

Take P(x) = a₀ + a₁x¹ + a₂x² + a₃x³ + ... + a_nxⁿ

If you assume z is a root, then sub z into the RHS and make the RHS = 0

a₀ + a₁z¹ + a₂z² + a₃z³ + ... + a_nzⁿ = 0

Then take the conjugate of both sides, manipulate using addition of conjugate rules and the fact that conjugates of real numbers are the same real number, and youll find you get z bar also being a root.