Complex Numbers  TOPIC_SOLVED

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

Complex Numbers

Postby lee on Wed Oct 02, 2013 7:52 pm

Hi

Can anyone help, been struggling for hours.


(i) given that w=1+2i, express w2, w3, and w4 in the form a+bi.

(ii) given that w is a root of the equation z4+pz3+qz2-6z+65=0, find the values of p and q.

(iii) write down a second root of the equation.

(iv) find the other 2 roots of the equation.

Thanks
lee
 
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Postby stapel_eliz on Wed Oct 02, 2013 8:00 pm

lee wrote:(i) given that w=1+2i, express w2, w3, and w4 in the form a+bi.

(ii) given that w is a root of the equation z4+pz3+qz2-6z+65=0, find the values of p and q.

The numbers which follow the variables: Are they exponents, so "w2" means "w^2"?

lee wrote:Can anyone help, been struggling for hours.

What have you tried? How far have you gotten? Please be complete (even if you think it's wrong), so the helpers can see what's going on. Thank you. :wink:
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Re: Complex Numbers

Postby lee on Wed Oct 02, 2013 8:10 pm

Yes i should have wrote it as w^2 etc

As for how far i have got, not very im afraid.

thanks
lee
 
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Re: Complex Numbers  TOPIC_SOLVED

Postby little_dragon on Wed Oct 02, 2013 9:48 pm

lee wrote:(i) given that w=1+2i, express w2, w3, and w4 in the form a+bi.

(ii) given that w is a root of the equation z4+pz3+qz2-6z+65=0, find the values of p and q.

(iii) write down a second root of the equation.

(iv) find the other 2 roots of the equation.

(i) do the powers: w^2 = (1+2i)^2 = (1+2i)(1+2i) = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i
do the same for the others
(ii) since 1+2i is a root then 1-2i is a root too
so do the synthetic div w/ each root: http://www.purplemath.com/modules/synthdiv.htm
put the remainder =0
i get (68-3q+5p)+(-4+4q-2p)i=0=0*1+0*i
this gives a linear system: 5p-3q=-68, -2p+4q=4
solve
(iii) since 1+2i is a root so is 1-2i
divide these out
this leaves a quadratic
solve

if U get stuck plz write back w/ your working
thnx
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Re: Complex Numbers

Postby lee on Fri Oct 04, 2013 6:16 pm

Thanks your answer really helped. :clap:
lee
 
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