Complex numbers (2 Viewers)

[dasicmankev]

Suppose that z^7 = 1 where z=/=1

(i) Deduce that z^3 + z^2 + z + 1 + 1/z + 1/z^2 + 1/z^3 = 0

(ii) By letting x = z + 1/z reduce the equation in (i) to a cubic equation in x.

(iii) Hence deduce that

(cos pi/7)(cos 2pi/7)(cos 3pi/7) = 1/8

I got (i) and (ii) but have no idea on how to start (iii). Any help would be greatly appreciated.

 

[ninetypercent]

z^7 = 1
z = cis0, cis 2pi/7, cis 4pi/7, cis6pi/7, ...etc

Find roots of the cubic equation using:
x = z + 1/z

then multiply all x's together, then equate with
product of roots = -d/a

 

[dasicmankev]

Thanks for the reply. I'm still a bit lost though, because the cubic equation is x^3 + x^2 - 2x - 1 = 0 (I think). So how do you find the roots to this equation by using x = z + 1/z? I mean there are 7 values for z.

 

[ninetypercent]

(ii) By letting x = z + 1/z reduce the equation in (i) to a cubic equation in x.

(z + 1/z)^3 - 3z - 3/z + (z + 1/z)^2 - 2 + (1/z + z) + 1= 0
x^3 - 3(z + 1/z) + (z+1/z)^2 - 2 + (z +1/z) + 1 = 0
x^3 - 3x + x^2 - 2 + x + 1= 0
x^3 + x^2 - 2x - 1 = 0

Your equation is correct

z = cis0, cis 2pi/7, cis 4pi/7, cis6pi/7, cis-2pi/7, cis-4pi/7, cis-6pi/7

for cis2pi/7
x = cis2pi/7 + 1/(cis2pi/7)
= 2cos(2pi/7)
x = 2Re(z)

so roots of cubic are 2cos 2pi/7, 2cos 4pi/7, 2cos6pi/7, 2cos2pi/7, 2cos4pi/7, 2cos6pi/7

multiply all of them

2cos 2pi/7x 2cos 4pi/7x2cos6pi/7 x 2cos2pi/7 x 2cos4pi/7x 2cos6pi/7 = 1 (as -d/a = 1)
64 cos^2(2pi/7) cos^2(4pi/7) cos^2 (6pi/7) = 1
cos^2(2pi/7) cos^2(4pi/7) cos^2 (6pi/7) = 1/64
taking the square root
cos(2pi/7) cos(4pi/7)cos(6pi/7) = 1/8
which is equal to:
(cos pi/7)(cos 2pi/7)(cos 3pi/7) = 1/8

 

[vafa]

View attachment 20177

View attachment 20178

 

[dasicmankev]

Thanks vafa, very clear working. But may I ask, how does (cos pi/7)(cos 2pi/7)(cos 3pi/7) equate to (cos 2pi/7)(cos 4pi/7)(cos 6pi/7) as ninetypercent stated? Now I know that both yield 1/8, but exactly can they be proved to be equivalent (other than just stating that their values are the same)?

 

[ninetypercent]

if you think back to the quadrants, cos is negative in the second quadrant

cos( pi - 6pi/7) is equal to cos(pi/7)
yet it is also equal to -cos(6pi/7)

also,
cos(pi - 4pi/7) = -cos(4pi/7) = cos(3pi/7)

 

[dasicmankev]

Noted. Thanks for clarifying.

 

[Lukybear]

I dont get how x1x2x3 = 1. Could someone explain?

 

[gurmies]

I dont get how x1x2x3 = 1. Could someone explain?

Product of roots is -d/a