How do you simplify $(sqrt3-i)div(2-2sqrt3i)$ ?

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Answer 1 · 2016-12-04T14:46:08

The answer is $=sqrt3/4+i/4$

When you have a fraction of complex numbers,

$w=z_1/z_2$

Multiply numerator and denominator by the conjugate of the denominator

$w=(z_1*barz_2)/(z_2*barz_2)$

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

The conjugate is $barz=a-ib$

and $i^2=-1$

$w=(sqrt3-i)/(2-2sqrt3i)$

$=((sqrt3-i)(2+2sqrt3i))/((2-2sqrt3i)(2+2sqrt3i))$

$=(2sqrt3+2*3i-2i-2sqrt3i^2)/(4-4*3i^2$

$=(4sqrt3+4i)/16$

$=sqrt3/4+i/4$

How do you simplify (sqrt3-i)div(2-2sqrt3i)(sqrt3-i)div(2-2sqrt3i)?