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

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Answer 1 · 2014-11-06T16:23:59

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

by rewriting in fraction form,

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

by factoring $2$ $2$ out of the denominator,

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

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

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

by simplifying a bit further,

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

by cancelling out $2$ $2$ ,

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

I hope that this was helpful.

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