How do you find the quotient of $(\sqrt{3}-i) -: (2- 2\sqrt{3}i)$ ?

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Answer 1 · 2014-11-06T16:23:59

$(sqrt{3}-i) divide (2-i2sqrt{3})$

by rewriting in fraction form,

$={sqrt{3}-i}/{2-2sqrt{3}i}$

by factoring $2$ out of the denominator,

$={sqrt{3}-i}/{2(1-sqrt{3}i)}$

by multiplying the numerator and the denominator by $(1+sqrt{3}i)$ ,

$={sqrt{3}+3i-i+sqrt{3}}/{2[1-(sqrt{3}i)^2]}$

by simplifying a bit further,

$={2(sqrt{3}+i)}/{2(1+3)}$

by cancelling out $2$ ,

$={sqrt{3}+i}/4$

I hope that this was helpful.

How do you find the quotient of (\sqrt{3}-i) -: (2- 2\sqrt{3}i)(\sqrt{3}-i) -: (2- 2\sqrt{3}i)?