How do you simplify $(\sqrt{3} - i) \div (2 - 2 \sqrt{3} i)$ $(sqrt3-i)div(2-2sqrt3i)$ ?

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Answer 1 · 2016-12-04T14:46:08

The answer is $= \frac{\sqrt{3}}{4} + \frac{i}{4}$ $=sqrt3/4+i/4$

When you have a fraction of complex numbers,

$w = \frac{z_{1}}{z_{2}}$ $w=z_1/z_2$

Multiply numerator and denominator by the conjugate of the denominator

$w = \frac{z_{1} \cdot {¯ z}_{2}}{z_{2} \cdot {¯ z}_{2}}$ $w=(z_1*barz_2)/(z_2*barz_2)$

If the complex number is $z = a + i b$ $z=a+ib$

The conjugate is $¯ z = a - i b$ $barz=a-ib$

and $i^{2} = - 1$ $i^2=-1$

$w = \frac{\sqrt{3} - i}{2 - 2 \sqrt{3} i}$ $w=(sqrt3-i)/(2-2sqrt3i)$

$= \frac{(\sqrt{3} - i) (2 + 2 \sqrt{3} i)}{(2 - 2 \sqrt{3} i) (2 + 2 \sqrt{3} i)}$ $=((sqrt3-i)(2+2sqrt3i))/((2-2sqrt3i)(2+2sqrt3i))$

$= \frac{2 \sqrt{3} + 2 \cdot 3 i - 2 i - 2 \sqrt{3} i^{2}}{4 - 4 \cdot 3 i^{2}}$ $=(2sqrt3+2*3i-2i-2sqrt3i^2)/(4-4*3i^2$

$= \frac{4 \sqrt{3} + 4 i}{16}$ $=(4sqrt3+4i)/16$

$= \frac{\sqrt{3}}{4} + \frac{i}{4}$ $=sqrt3/4+i/4$

How do you simplify (sqrt3-i)div(2-2sqrt3i)(√3−i)÷(2−2√3i)(sqrt3-i)div(2-2sqrt3i)?