little help... (1 Viewer)

[routemarker]

can anyone here show the working for the questions below and how to do them

1) express all solutions of z^6-64 as a product of real linear/irreducible quadratic factors

2) express z^5 +1 as a product of linear factors. complex numbers can be left in polar form.

cheers

 

[McLake]

Originally posted by routemarker
can anyone here show the working for the questions below and how to do them

1) express all solutions of z^6-64 as a product of real linear/irreducible quadratic factors

2) express z^5 +1 as a product of linear factors. complex numbers can be left in polar form.

cheers

Are you doing the MATH1131/1141 algebra test tommorow? I think so ...

1)
z^6 = 64
|z| = 2
6@ = 0 , +/- 2pi, +/- 4pi, +/-6pi
@ = 0, +/-(pi/3), +/-(2pi/3), pi
so solutions are 2, 2e^(pi*i/3), 2e^(-pi*i/3) ...

and then factorise:

(z +/- pi)(z^2 + 4cos(2pi/3) + 4)(z^2 + 4cos(4pi/3) + 4)


2)
z^5 = -1
|z| = 1
5@ = +/- pi, +/- 2pi, +/- 4pi
@ = +/-(pi/5), +/-(2pi/5), (4pi/5)
so solutions are e^(pi*i/5) ....

and factorise:

(z - e^(4pi*i/5))(z^2 + 2cos(pi/5) + 1)(z^2 + 2cos(2pi/5) + 1)

 

[spice girl]

MATH1141....NOOOO....

 

[ND]

Re: Re: little help...

Originally posted by McLake


Are you doing the MATH1131/1141 algebra test tommorow? I think so ...

1)
z^6 = 64
|z| = 2
6@ = 0 , +/- 2pi, +/- 4pi, +/-6pi
@ = 0, +/-(pi/3), +/-(2pi/3), pi
so solutions are 2, 2e^(pi*i/3), 2e^(-pi*i/3) ...

and then factorise:

(z +/- pi)(z^2 + 4cos(2pi/3) + 4)(z^2 + 4cos(4pi/3) + 4)


2)
z^5 = -1
|z| = 1
5@ = +/- pi, +/- 2pi, +/- 4pi
@ = +/-(pi/5), +/-(2pi/5), (4pi/5)
so solutions are e^(pi*i/5) ....

and factorise:

(z - e^(4pi*i/5))(z^2 + 2cos(pi/5) + 1)(z^2 + 2cos(2pi/5) + 1)

Just a q: Why did you use euler's identity to convert the soln's to exponents?

 

[McLake]

Re: Re: Re: little help...

Originally posted by ND


Just a q: Why did you use euler's identity to convert the soln's to exponents?

Because that's how we do it at uni (and I'm sure routemarker was asking these as uni questions)

 

[routemarker]

thanks mclake, you made it much clearer.

 

[maniacguy]

in this case, easier is:
z^6 - 64
= (z^3 - 8)(z^3 + 8)
= (z-2)(z^2 + 2z+4)(z+2)(z^2-2z+4)

for the first one.

If you can see difference of squares / cubes factorization, then use it!