complex factorising qn (1 Viewer)

[Masaken]

i approached the qn in a way that was wrong, though i don't understand what they're doing here, can someone explain please? thanks in advance

 

[Life'sHard]

In layman’s terms. Factorise z^5+1 into quadratics that cannot be simplified further and that the coefficients of z are real. The general factoring form of something to the power of 5 is given in the first line. From part 1 you’ve solved for the roots which can easily be subbed in. They’ve now grouped each factor together so that when u expand it, the isin gets canceled out and you’re left with a real coefficient of cos. Then you can stop after expanding.

 

[ImagineDragin]

In layman’s terms. Factorise z^5+1 into quadratics that cannot be simplified further and that the coefficients of z are real. The general factoring form of something to the power of 5 is given in the first line. From part 1 you’ve solved for the roots which can easily be subbed in. They’ve now grouped each factor together so that when u expand it, the isin gets canceled out and you’re left with a real coefficient of cos. Then you can stop after expanding.

How do you figure out which z to group together to cancel out the i? So say your in an exam, how would you figure it out?

 

[Masaken]

In layman’s terms. Factorise z^5+1 into quadratics that cannot be simplified further and that the coefficients of z are real. The general factoring form of something to the power of 5 is given in the first line. From part 1 you’ve solved for the roots which can easily be subbed in. They’ve now grouped each factor together so that when u expand it, the isin gets canceled out and you’re left with a real coefficient of cos. Then you can stop after expanding.

when putting in the roots i just discern that they're a difference of two squares then solve from there, right? because the roots are conjugates of each other i can do z^2 - 2zRe(root) + 1 ?

 

[Drongoski]

Another way of looking at the situation is that the equation z^5 - 1 = 0 has 5 roots, a, b, c, d and e say.
In this case z^5 - 1 = (z-a)(z-b)(z-c)(z-d)(z-e). For z^5 - 1 = 0, we have 5 roots, one real and the other 4 being complex.
Since z^5 - 1 = 0 is a polynomial eqn with real coefficients, complex roots must occur in conjugate pairs.



= (1 real linear factor) x (1 real quadratic factor) x (1 real quadratic factor)

Oh dear, I got the wrong equation: z^5 - 1 = 0 instead of z^5 + 1 = 0. Should have gone to Specsavers!

 

[Life'sHard]

How do you figure out which z to group together to cancel out the i? So say your in an exam, how would you figure it out?

So yeah like drongoski said. Summing conjugate roots should make the result real.

 

[Drongoski]

So yeah like drongoski said. Summing conjugate roots should make the result real.

You can see from the simple, real, quadratic equation: