complex number exponentials

**mistykz** · September 10th 2007, 05:36 PM

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?

**Plato** · September 10th 2007, 06:13 PM

You must learn to find the arguments of complex numbers!
$\arg \left( {\sqrt 3 - 3i} \right) = \arctan \left( {\frac{{ - 3}}{{\sqrt 3 }}} \right) = \arctan \left( { - \sqrt 3 } \right) = - \frac{\pi }{3}$ .

Now all the three questions depend upon two facts: $\arg \left( {\sqrt 3 - 3i} \right) = - \frac{\pi }{3}$ and $\left| {\sqrt 3 - 3i} \right| = \sqrt {12} = 2\sqrt{3}$

**mistykz** · September 10th 2007, 06:40 PM

okay, so now what?

**Jhevon** · September 10th 2007, 06:59 PM

Originally Posted by mistykz

okay, so now what?

we can express the complex number $x + iy$ as $r e^{i \theta}$ , where $r = |x + iy| = \sqrt {x^2 + y^2}$ and $\theta = \arg (x + iy)$

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.

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