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

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

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Hint: Was the term on the left supposed to equal the term on the right or are you supposed to simplify the entire expression? Anyway, you can use DMT on the denominator first, and then simplify the remaining fraction (the numerator and denominator) and then clean up whatever remains. Is that clear? Good luck! –  Amzoti Dec 21 '12 at 22:47
    
In the problem it is on the denominator ((1-isquareroot(3))^3 they are not separated. –  Fernando Martinez Dec 21 '12 at 22:50
    
Also on the numerator it is (1+i)(square root3+i)^3 I am not sure if I forgot to add the 3 exponent. –  Fernando Martinez Dec 21 '12 at 22:53
    
Please help ayuda. I am unsure how to solve I tried multiplying out the numerator but I still have issues. –  Fernando Martinez Dec 21 '12 at 22:54
    
Give me a sec to correct the problem. Is the problem correct now? –  Amzoti Dec 21 '12 at 22:57
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1 Answer

up vote 2 down vote accepted

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

$ \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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Very good answer. –  Fernando Martinez Dec 22 '12 at 0:17
    
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. –  user50407 Dec 22 '12 at 0:22
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