Complex Numbers

[paotra]

If a polynomial with real coefficients has solutions z=1-ai and z=2 over the set of complex numbers, where a is real, then its quadratic factor must be?

a. z^2 + 4
b. z^2 - 2aiz + 1
c. z^2 - 2z + 1 + a^2
d. z^2 - 4
e. z^2 + 2z + 1 - a^2

How come we just can't simply do (z-1+ai)(z-2) ? isn't that also a quadratic?

All suggestions are welcome, thanks

[vea]

How can it only have a soln of ?
Should there be it's conjugate of as well?

If yes, then the quadratic factor should be





which is C

[paotra]

It is C, however how come (z-1+ai)(z-2) is not considered? Is it only because in this case the answers do not have (z-1+ai)(z-2)?

[vea]

I think they also want the quadratic factor to be real- I'm not too sure about this though.

[jimmy999]

It is C, however how come (z-1+ai)(z-2) is not considered? Is it only because in this case the answers do not have (z-1+ai)(z-2)?

You are correct in thinking that is a quadratic factor. In fact there are 3 different quadratic factors, it just depends on which linear factors you choose. For this question, only one of them is a quadratic factor so you have to pick C.

I think they also want the quadratic factor to be real- I'm not too sure about this though.

If the solutions are in complex conjugates, then the quadratic factor must be real as the conjugate root theorem only holds for real coefficients