complex roots, general solutions (1 Viewer)

vds700 · Mar 14, 2009

hey guys, i got this question from a uni assignment on complex numbers, and i cant remember how to find complex roots in polar form.

OK I get

vds700 · Mar 14, 2009

$cos3\theta =cos\alpha \\$Now the general solution is 3\theta =2k\pi \pm \alpha right??$\\$So$ \theta =\frac{2k\pi \pm \alpha }{3}\\$If I were to sub k = 0, \pm 1, would I get the required solutions?$$

Thanks guys

Drongoski · Mar 14, 2009

vds700 said:
$cos3\theta =cos\alpha \\$Now the general solution is 3\theta =2k\pi \pm \alpha right??$\\$So$ \theta =\frac{2k\pi \pm \alpha }{3}\\$If I were to sub k = 0, \pm 1, would I get the required solutions?$$

Thanks guys

Seems correct to me.

vds700 · Mar 14, 2009

Drongoski said:
Seems correct to me.

im just a bit confused coz the lecturer had general solution 2kpi+ alpha when he id an example, but in my HSC notes, ive got 2kpi + alpha

Does it matter what values of k you sub in?

radio-waves · Mar 14, 2009

No it doesn't matter, the values of k think of them as revolutions.

So its perfectly fine with 2kpi +- alpha

vds700 · Mar 14, 2009

radio-waves said:
No it doesn't matter, the values of k think of them as revolutions.

So its perfectly fine with 2kpi +- alpha

ok thanks

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complex roots, general solutions (1 Viewer)

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