Complex numbers- quickques. (1 Viewer)

[skillz]

Factorise the first expression into linear factors over c, given that the second expression is one of the linear factors.

z^3+(1-i)z^2+(1-i)z-i, z- i


can someone please solve that for me?i'm a bit confused with ho/w to do it..

thanks alot

 

[Mill]

p(z) = z³ + (1 - i)z² + (1 - i)z - i
= (z - i)(z² + z + 1) by inspection

You can now factorise that last quadratic yourself.

Alternatively, if you aren't comfortable with 'by inspection' you could:

1. Use long division

2. Rearrange the polynomial into a more convenient form for factorsation by 'grouping in pairs' (except it isn't pairs).

ie. p(z) = z³ + (1 - i)z² + (1 - i)z - i
= z³ + z² + z - iz² - iz - i
= z(z² + z + 1) - i(z² + z + 1)
= (z - i)(z² + z + 1)