Please help me answer both of these questions. Thank you :)

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- Jun 15th 2006, 10:03 PMnoriveaComplex numbers
Please help me answer both of these questions. Thank you :)

- Jun 15th 2006, 11:47 PMCaptainBlackQuote:

Originally Posted by**norivea**

$\displaystyle

(z-3)(z^2-5z+8)=0

$

This has three roots, one real corresponding to the first factor; so $\displaystyle x=3$ is a root.

The other two roots are the roots of the second factor $\displaystyle z^2-5z+8=0$, these may

be found using the quadratic formula, which gives in this case:

$\displaystyle

z=\frac{5}{2}\pm \frac{\sqrt{7}}{2}i

$

RonL - Jun 16th 2006, 12:39 AMCaptainBlackQuote:

Originally Posted by**norivea**

$\displaystyle

z^3-10z^2+34z-40

$

given that $\displaystyle 3-i$ is a root.

A cubic with real coefficients always has at least one real root, and

complex roots occur in conjugate pairs. So both $\displaystyle 3-i$ and

$\displaystyle 3+i$ are roots. Also:

$\displaystyle

z^3-10z^2+34z-40=(z-(3-i))(z-(3+i))(z-a)

$

where $\displaystyle a$ is real, and so easily found to be $\displaystyle 4$,

so the remaining root is $\displaystyle z=4$.

RonL