simplify, in cartesian and polar forms:

(sqrt(3)-3i)^10

(sqrt(3)-3i)^ -1

(sqrt(3)-3i)^ 1/3

can someone please explain how i would go about doing this?

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- Sep 10th 2007, 05:36 PMmistykzcomplex number exponentials
simplify, in cartesian and polar forms:

(sqrt(3)-3i)^10

(sqrt(3)-3i)^ -1

(sqrt(3)-3i)^ 1/3

can someone please explain how i would go about doing this? - Sep 10th 2007, 06:13 PMPlato
You must learn to find the arguments of complex numbers!

$\displaystyle \arg \left( {\sqrt 3 - 3i} \right) = \arctan \left( {\frac{{ - 3}}{{\sqrt 3 }}} \right) = \arctan \left( { - \sqrt 3 } \right) = - \frac{\pi }{3}$.

Now all the three questions depend upon two facts: $\displaystyle \arg \left( {\sqrt 3 - 3i} \right) = - \frac{\pi }{3}$ and $\displaystyle \left| {\sqrt 3 - 3i} \right| = \sqrt {12} = 2\sqrt{3} $ - Sep 10th 2007, 06:40 PMmistykz
okay, so now what?

- Sep 10th 2007, 06:59 PMJhevon
we can express the complex number $\displaystyle x + iy$ as $\displaystyle r e^{i \theta}$, where $\displaystyle r = |x + iy| = \sqrt {x^2 + y^2}$ and $\displaystyle \theta = \arg (x + iy)$

in this form, it is easy to apply the powers and simplify

you can see the "Background" section in the first post here for more information.