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$\begingroup$

$\frac{(1+i)(\sqrt{3} + i)^3}{(1-\sqrt{3}i)^{3}} = 1-i$

What confuses me is how would I do the numerator because I have two expressions.

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    yes it is correct2012-12-21

1 Answers 1

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$1+i=\sqrt 2 (\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$

$ \sqrt3+i=2(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})) $

$ 1-i\sqrt3=2(\cos(\frac{-\pi}{3})+i\sin(\frac{-\pi}{3})$

Then you can simply apply De Moivre's theorem:

The numerator becomes $8\sqrt2(\cos(\frac{9\pi}{12})+i\sin(\frac{9\pi}{12}))=8\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))$

The denominator becomes $8(\cos(-\pi)+i\sin(-\pi))=-8$

So the fraction is equal to $-\sqrt2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=1-i$

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    Thank you, a good thing to keep in mind with this kind of question (I think) is the unit circle and how $\sin$ and $\cos$ appear in it. Then it shouldn't be too much of a problem to find the arguments and moduli of the terms and apply De Moivre's Theorem.2012-12-22