Complex numbers (1 Viewer)

[yeyjasminetime]

Hello there!

Could someone explain what I means when you 'find 3 cube roots of each of the following in the form a + ib?

a) 1
b) -1

 

[HeroicPandas]

a) z³ = 1

b) z³ = -1

 

[yeyjasminetime]

Thanks

Z^3=-i^2?

 

[HeroicPandas]

Thanks

Z^3=-i^2?

Dont bring in powers of TWO because since u already have a pwoer of THREE, and a 1, u can use cubic factorisation

If u did it ur way:
z^3 = -i^2
z^3 + i^2 = 0

How do u solve this?

Go back,
z^3 = -i^2

If u REALLY wanna keep the i², then let z = rcis(theta) to solve for 'z'

 

[yeyjasminetime]

Ohhhhhhh! Yes I understand now
Z^3-1^3

 

[hit patel]

Well ok
Hmmm.
z^3=1
therefore z^3=cis 0
z= cube rt(cis 0)
z= cube rt (cis 0 + 2k pi)
Using De Moivre
z= cis (2k pi)/3
Let k = 0,+_1
z= 1, cis 2pi/3 , cis -2pi/3
z= 1, -1/2 + (sqrt(3)i)/2 , -1/2 - (sqrt(3)i)/2

Use a similar process for z^3+1=0 U can use this for other powers too like z^9+1.


Or you could factorise it for lower the lower powers like the one u provided and then use quadratic formula for the part of factored form to find ans then simplify it to get it in form a+ib

 

[MrBeefJerky]

b) z^3+1=0
(z+1)(z^2-z+1)=0
discriminant: -3
z= 1/2 +- sqr(3)i/2
Therefore z = -1, 1/2 +- sqr3i/2