maths question - complex numbers

I've tried to do this about 5 times now and I'm getting really frustrated, if anybody could help me, I'd really appreciate it:

Z=2 - j2 is a root of 2z^3 - 9z^2 + 20z -8 = 0
find the remainding roots of the equation:

I'm thinking 2 + j2 would also be a root because it's the complex conjugate, if that's correct?

Thanks guys!! x

Reply 1

nota bene

15

Yes, the conjugate is a root, so one left to find, should be possible by factorisation.

Reply 2

PilotClaire

OP

Thanks, but I'm still a bit lost.
What would I factorise?

I tried:

2z^3 - 9z^2 + 20z - 8 = (2-j2)(2+j2)(x+jy)

and tried to find out what the values of x and y would be, but that didn't work.

Reply 3

insparato

15

if 2-j2 and 2+j2 are factors

(z - (2-j2))(z-(2+j2))
is also a factor
(z^2 - 4z + 8)
therefore

2z^3 - 9z^2 + 20z - 8 = (z^2 - 4z + 8)(az + b)

compare coefficients of z^3

az^3 = 2z^3

a = 2

compare coefficients of numbers

8b = -8

b = -1

So

2z^3 - 9z^2 + 20z - 8 = (z^2 - 4z + 8)(2z - 1)

z = 1/2
z = 2-j2
z = 2+j2

Reply 4

PilotClaire

OP

Thank you so much.

Reply 5

mrsmann

2

Sorry to hijack the thread but there's no point starting another one.

I have a question for roots. If you have a quadratic whose solutions are complex, then complex conjugates are roots because in the quadratic formulae there's (+/-) square root of b^2 - 4ac. So this proves that for quadratics complex conjugates are roots. But for any higher power what is the proof? Is it sufficient to assume that because the function probably factorises into a quadratic and a (ax+b) (for a function with highest term x^3) and so obviously it must have conjugate roots from the quadratic? Is that it?

Reply 6

nota bene

15

mrsmann

I have a question for roots. If you have a quadratic whose solutions are complex, then complex conjugates are roots because in the quadratic formulae there's (+/-) square root of b^2 - 4ac. So this proves that for quadratics complex conjugates are roots. But for any higher power what is the proof? Is it sufficient to assume that because the function probably factorises into a quadratic and a (ax+b) (for a function with highest term x^3) and so obviously it must have conjugate roots from the quadratic? Is that it?

There is no need for me to copy the wiki article, so here you go :p:

LINK

Reply 7

generalebriety

16

mrsmann

Sorry to hijack the thread but there's no point starting another one.

I have a question for roots. If you have a quadratic whose solutions are complex, then complex conjugates are roots because in the quadratic formulae there's (+/-) square root of b^2 - 4ac. So this proves that for quadratics complex conjugates are roots. But for any higher power what is the proof? Is it sufficient to assume that because the function probably factorises into a quadratic and a (ax+b) (for a function with highest term x^3) and so obviously it must have conjugate roots from the quadratic? Is that it?

The "fundamental theorem of algebra" states that every polynomial equation

k_0 + k_1x + k_2x^2 + \dots + k_nx^n = 0

can be factorised into

(x - x_0)(x - x_1)(x - x_2)\dots (x-x_n)

, where the

x_i \in \mathbb{C}

are the roots of the equation. If the

k_i

are all real then there can't be a complex

x_i

without

x_j = x_i ^*

for some j, otherwise there'd be complex coefficients.

Note that if the

k_i

are complex this doesn't hold...

Reply 8

mrsmann

2

Thanks for the replies!! nota bene that "leisurely" proof is harder to understand than general's!!!

But thanks

maths question - complex numbers

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