complex numbers question (1 Viewer)

[elseany]

straight from cambridge P74 Q12:

show that the roots of z^6 + z^3 + 1 = 0 are among the roots of z^9 -1 = 0. hence find the roots of z^6 + z^3 + 1 = 0 in modulus/argument form.

 

[Trebla]

Using the differences of two cubes formula:
z⁹ - 1 = (z³ - 1)(z⁶ + z³ + 1)
Since z⁹ - 1 = 0
.: (z³ - 1)(z⁶ + z³ + 1) = 0
.: (z⁶ + z³ + 1) is among the roots of z⁹ - 1 = 0

To find roots of (z⁶ + z³ + 1) = 0, we can find roots of z⁹ - 1 = 0 where z³ =/= 1 (i.e. z =/= 1, cis 2π/3, cis 4π/3)

.: z = cis (2π/9), cis (4π/9), cis (8π/9), cis (10π/9), cis (14π/9), cis (16π/9)
i.e. z = cis (±2π/9), cis (±4π/9), cis (±8π/9)

 

[jyu]

straight from cambridge P74 Q12:

show that the roots of z^6 + z^3 + 1 = 0 are among the roots of z^9 -1 = 0. hence find the roots of z^6 + z^3 + 1 = 0 in modulus/argument form.

z^9 -1 = (z^3 - 1)(z^6 + z^3 + 1)

.: the roots of z^6 + z^3 + 1 = 0 are among the roots of z^9 -1 = 0.

The roots of z^9 -1 = 0 are 1cis(2npi/9) where n = 0, 1, 2, 3, ......, 8 or
n = 0, +/-1, +/-2, +/-3, +/-4.

.: the roots of z^6 + z^3 + 1 = 0 are 1cis(2npi/9) where n = 1, 2, 4, 5, 7, 8 or
n = +/-1, +/-2, +/-4.

:wave:

 

[elseany]

thankks for that, havnt learnt that expansion yet ><

but how do u know that all the roots of the bigger polynomial are going to be roots of the smaller one?

for example y=(x-1)(x+2) roots of y are 1 and -2, yet the roots of y=x-1 is just 1.

get what i mean? >_< sorry if this is confusing

 

[Trebla]

thankks for that, havnt learnt that expansion yet ><

but how do u know that all the roots of the bigger polynomial are going to be roots of the smaller one?

for example y=(x-1)(x+2) roots of y are 1 and -2, yet the roots of y=x-1 is just 1.

get what i mean? >_< sorry if this is confusing

From (z³ - 1)(z⁶ + z³ + 1) = 0
(z⁶ + z³ + 1) = 0, only if (z³ - 1) =/= 0 (if it does equal zero then we would be dividing by zero which we cannot do at this stage)
So for (z³ - 1) =/= 0, this means that the roots of (z³ - 1) cannot be included in the final solution.

Did that answer your question?

 

[elseany]

ahhh excellent!

yes that clears it up thanks

btw did u change ur post or did i completely miss the z^3 =/= 1 part? ><