http://i49.tinypic.com/2m7hnro.png

However, this is wrong. It should be $\displaystyle z^2+zi-1$. I'm not sure how one would arrive here, however.

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- Jan 25th 2010, 06:21 PMdavismj[SOLVED] Complex Polynomial Division
http://i49.tinypic.com/2m7hnro.png

However, this is wrong. It should be $\displaystyle z^2+zi-1$. I'm not sure how one would arrive here, however. - Jan 25th 2010, 06:23 PMdavismj
Wait, nevermind. I see it now. Stupid mistake. Sorry.

- Jan 25th 2010, 07:09 PMProve It
Try using the Difference of Two Cubes Rule:

$\displaystyle z^3 + i = z^3 - (-i)$

$\displaystyle = z^3 - i^3$

$\displaystyle = (z - i)(z^2 + iz + i^2)$

$\displaystyle = (z - i)(z^2 + iz - 1)$.

So $\displaystyle \frac{z^3 + i}{z - i} = \frac{(z - i)(z^2 + iz - 1)}{z - i}$

$\displaystyle = z^2 + iz - 1$.