Roots of a squared polynomial ( complex numbers)

[Axe199]

Homework Statement

problem in a pic attached

Homework Equations

The Attempt at a Solution

i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i don't know what he wants from me

 

[Ray Vickson]

Homework Statement

problem in a pic attached

Homework Equations

The Attempt at a Solution

i solved i and ii a , when it came to b , i just said that every one of the 3 roots will be squared having 2 roots 1 + and 1 - but then i read the marking schemes ( also attached) , and i got really confused and now i don't know what he wants from me

He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.

 

[Axe199]

He wants you to tell him what are all the square roots of -2, -2+2*sqrt(3)*i and -2 - 2*sqrt(3)*i.

why can't i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?

 

[slider142]

why can't i just say (z+2)^2= 0 and solve and do the same for the complex roots?
why do i have to find the roots of the -2?

(z + 2)² is not a factor of p(z²). It is a factor of (p(z))², however, a completely different polynomial.
The polynomial is not the object being squared. p(z²) only means that the variable z is being replaced with the variable z². So, as a result, we know that z² + 2 is a factor.

 

[Axe199]

(z + 2)² is not a factor of p(z²). It is a factor of (p(z))², however, a completely different polynomial.
The polynomial is not the object being squared. p(z²) only means that the variable z is being replaced with the variable z². So, as a result, we know that z² + 2 is a factor.

okay...thanks very much, tht was a tricky one, or maybe i am stupid :D
either way , thanks , i got it